从 k 个列表中查找包含元素的最小范围

给定 k 个大小为 n 的整数的排序列表，从 k 个列表中的每个列表中找到至少包含元素的最小范围。如果找到多个最小范围，则打印其中任何一个。

例子：

Input: K = 3
arr1[] : [4, 7, 9, 12, 15]
arr2[] : [0, 8, 10, 14, 20]
arr3[] : [6, 12, 16, 30, 50]
Output:
The smallest range is [6 8]

Explanation: Smallest range is formed by 
number 7 from the first list, 8 from second
list and 6 from the third list.

Input: k = 3
arr1[] : [4, 7]
arr2[] : [1, 2]
arr3[] : [20, 40]
Output:
The smallest range is [2 20]

Explanation:The range [2, 20] contains 2, 4, 7, 20
which contains element from all the three arrays.

朴素方法：给定 K 个排序列表，找到一个范围，其中每个列表中至少有一个元素。解决这个问题的思路很简单，保持k个指针构成范围内的元素，取k个元素的最小值和最大值就可以形成范围。最初，所有指针都将指向所有 k 个数组的开头。存储范围最大值到最小值。如果必须最小化范围，则必须增加最小值或必须减小最大值。数组排序后，最大值不能减少，但可以增加最小值。要继续增加最小值，请增加包含最小值的列表的指针并更新范围，直到列表之一耗尽。

• 算法：
  1. 创建一个长度为 k 的额外空间ptr来存储指针和一个初始化为最大值的变量minrange 。
  2. 最初每个列表的索引都是 0，因此将ptr[0..k] 的每个元素初始化为 0，数组 ptr 将存储范围内元素的索引。
  3. 重复以下步骤，直到至少一个列表用完：
     1. 现在找到 ptr[0…k] 数组指向的所有列表的当前元素中的最小值和最大值。
     2. 如果当前 (max-min) 小于 minrange，则更新 minrange。
     3. 增加指向当前最小元素的指针。
• 执行：

C++

// C++ program to finds out smallest range that includes
// elements from each of the given sorted lists.
#include 
 
using namespace std;
 
#define N 5
 
// array for storing the current index of list i
int ptr[501];
 
// This function takes an k sorted lists in the form of
// 2D array as an argument. It finds out smallest range
// that includes elements from each of the k lists.
void findSmallestRange(int arr[][N], int n, int k)
{
    int i, minval, maxval, minrange, minel, maxel, flag, minind;
 
    // initializing to 0 index;
    for (i = 0; i <= k; i++)
        ptr[i] = 0;
 
    minrange = INT_MAX;
 
    while (1) {
        // for maintaining the index of list containing the minimum element
        minind = -1;
        minval = INT_MAX;
        maxval = INT_MIN;
        flag = 0;
 
        // iterating over all the list
        for (i = 0; i < k; i++) {
            // if every element of list[i] is traversed then break the loop
            if (ptr[i] == n) {
                flag = 1;
                break;
            }
            // find minimum value among all the list elements pointing by the ptr[] array
            if (ptr[i] < n && arr[i][ptr[i]] < minval) {
                minind = i; // update the index of the list
                minval = arr[i][ptr[i]];
            }
            // find maximum value among all the list elements pointing by the ptr[] array
            if (ptr[i] < n && arr[i][ptr[i]] > maxval) {
                maxval = arr[i][ptr[i]];
            }
        }
 
        // if any list exhaust we will not get any better answer, so break the while loop
        if (flag)
            break;
 
        ptr[minind]++;
 
        // updating the minrange
        if ((maxval - minval) < minrange) {
            minel = minval;
            maxel = maxval;
            minrange = maxel - minel;
        }
    }
 
    printf("The smallest range is [%d, %d]\n", minel, maxel);
}
 
// Driver program to test above function
int main()
{
    int arr[][N] = {
        { 4, 7, 9, 12, 15 },
        { 0, 8, 10, 14, 20 },
        { 6, 12, 16, 30, 50 }
    };
 
    int k = sizeof(arr) / sizeof(arr[0]);
 
    findSmallestRange(arr, N, k);
 
    return 0;
}
// This code is contributed by Aditya Krishna Namdeo

Java

// Java program to finds out smallest range that includes
// elements from each of the given sorted lists.
class GFG {
 
    static final int N = 5;
 
    // array for storing the current index of list i
    static int ptr[] = new int[501];
 
    // This function takes an k sorted lists in the form of
    // 2D array as an argument. It finds out smallest range
    // that includes elements from each of the k lists.
    static void findSmallestRange(int arr[][], int n, int k)
    {
        int i, minval, maxval, minrange, minel = 0, maxel = 0, flag, minind;
 
        // initializing to 0 index;
        for (i = 0; i <= k; i++) {
            ptr[i] = 0;
        }
 
        minrange = Integer.MAX_VALUE;
 
        while (true) {
            // for maintaining the index of list containing the minimum element
            minind = -1;
            minval = Integer.MAX_VALUE;
            maxval = Integer.MIN_VALUE;
            flag = 0;
 
            // iterating over all the list
            for (i = 0; i < k; i++) {
                // if every element of list[i] is traversed then break the loop
                if (ptr[i] == n) {
                    flag = 1;
                    break;
                }
                // find minimum value among all the list elements pointing by the ptr[] array
                if (ptr[i] < n && arr[i][ptr[i]] < minval) {
                    minind = i; // update the index of the list
                    minval = arr[i][ptr[i]];
                }
                // find maximum value among all the list elements pointing by the ptr[] array
                if (ptr[i] < n && arr[i][ptr[i]] > maxval) {
                    maxval = arr[i][ptr[i]];
                }
            }
 
            // if any list exhaust we will not get any better answer, so break the while loop
            if (flag == 1) {
                break;
            }
 
            ptr[minind]++;
 
            // updating the minrange
            if ((maxval - minval) < minrange) {
                minel = minval;
                maxel = maxval;
                minrange = maxel - minel;
            }
        }
        System.out.printf("The smallest range is [%d, %d]\n", minel, maxel);
    }
 
    // Driver program to test above function
    public static void main(String[] args)
    {
 
        int arr[][] = {
            { 4, 7, 9, 12, 15 },
            { 0, 8, 10, 14, 20 },
            { 6, 12, 16, 30, 50 }
        };
 
        int k = arr.length;
 
        findSmallestRange(arr, N, k);
    }
}
// this code contributed by Rajput-Ji

Python

# Python3 program to finds out
# smallest range that includes
# elements from each of the
# given sorted lists.
 
N = 5
 
# array for storing the
# current index of list i
ptr = [0 for i in range(501)]
 
# This function takes an k sorted
# lists in the form of 2D array as
# an argument. It finds out smallest
# range that includes elements from
# each of the k lists.
def findSmallestRange(arr, n, k):
 
    i, minval, maxval, minrange, minel, maxel, flag, minind = 0, 0, 0, 0, 0, 0, 0, 0
         
    # initializing to 0 index
    for i in range(k + 1):
        ptr[i] = 0
 
    minrange = 10**9
         
    while(1):   
         
            # for maintaining the index of list
            # containing the minimum element
        minind = -1
        minval = 10**9
        maxval = -10**9
        flag = 0
 
        # iterating over all the list
        for i in range(k):
             
                # if every element of list[i] is
                # traversed then break the loop
            if(ptr[i] == n):
                flag = 1   
                break
 
            # find minimum value among all the list
            # elements pointing by the ptr[] array
            if(ptr[i] < n and arr[i][ptr[i]] < minval):
                minind = i # update the index of the list
                minval = arr[i][ptr[i]]
             
            # find maximum value among all the
            # list elements pointing by the ptr[] array
            if(ptr[i] < n and arr[i][ptr[i]] > maxval):
                maxval = arr[i][ptr[i]]
             
         
 
        # if any list exhaust we will
        # not get any better answer,
        # so break the while loop
        if(flag):
            break
 
        ptr[minind] += 1
 
        # updating the minrange
        if((maxval-minval) < minrange):
            minel = minval
            maxel = maxval
            minrange = maxel - minel
     
    print("The smallest range is [", minel, maxel, "]")
 
# Driver code
arr = [
    [4, 7, 9, 12, 15],
    [0, 8, 10, 14, 20],
    [6, 12, 16, 30, 50]
    ]
 
k = len(arr)
 
findSmallestRange(arr, N, k)
 
# This code is contributed by mohit kumar

C#

// C# program to finds out smallest
// range that includes elements from
// each of the given sorted lists.
using System;
 
class GFG {
 
    static int N = 5;
 
    // array for storing the current index of list i
    static int[] ptr = new int[501];
 
    // This function takes an k sorted
    // lists in the form of 2D array as
    // an argument. It finds out smallest range
    // that includes elements from each of the k lists.
    static void findSmallestRange(int[, ] arr,
                                  int n, int k)
    {
        int i, minval, maxval, minrange,
            minel = 0, maxel = 0, flag, minind;
 
        // initializing to 0 index;
        for (i = 0; i <= k; i++) {
            ptr[i] = 0;
        }
 
        minrange = int.MaxValue;
 
        while (true) {
            // for maintining the index of
            // list containing the minimum element
            minind = -1;
            minval = int.MaxValue;
            maxval = int.MinValue;
            flag = 0;
 
            // iterating over all the list
            for (i = 0; i < k; i++) {
                // if every element of list[i]
                // is traversed then break the loop
                if (ptr[i] == n) {
                    flag = 1;
                    break;
                }
 
                // find minimum value among all the
                // list elements pointing by the ptr[] array
                if (ptr[i] < n && arr[i, ptr[i]] < minval) {
                    minind = i; // update the index of the list
                    minval = arr[i, ptr[i]];
                }
 
                // find maximum value among all the
                // list elements pointing by the ptr[] array
                if (ptr[i] < n && arr[i, ptr[i]] > maxval) {
                    maxval = arr[i, ptr[i]];
                }
            }
 
            // if any list exhaust we will
            // not get any better answer,
            // so break the while loop
            if (flag == 1) {
                break;
            }
 
            ptr[minind]++;
 
            // updating the minrange
            if ((maxval - minval) < minrange) {
                minel = minval;
                maxel = maxval;
                minrange = maxel - minel;
            }
        }
        Console.WriteLine("The smallest range is"
                              + "[{0}, {1}]\n",
                          minel, maxel);
    }
 
    // Driver code
    public static void Main(String[] args)
    {
 
        int[, ] arr = {
            { 4, 7, 9, 12, 15 },
            { 0, 8, 10, 14, 20 },
            { 6, 12, 16, 30, 50 }
        };
 
        int k = arr.GetLength(0);
 
        findSmallestRange(arr, N, k);
    }
}
 
// This code has been contributed by 29AjayKumar

Javascript

 
 
// Javascript program to finds out smallest range that includes
// elements from each of the given sorted lists.
let N = 5;
 
// array for storing the current index of list i
let ptr=new Array(501);
 
// This function takes an k sorted lists in the form of
    // 2D array as an argument. It finds out smallest range
    // that includes elements from each of the k lists.
function findSmallestRange(arr,n,k)
{
    let i, minval, maxval, minrange, minel = 0, maxel = 0, flag, minind;
   
        // initializing to 0 index;
        for (i = 0; i <= k; i++) {
            ptr[i] = 0;
        }
   
        minrange = Number.MAX_VALUE;
   
        while (true) {
            // for maintining the index of list containing the minimum element
            minind = -1;
            minval = Number.MAX_VALUE;
            maxval = Number.MIN_VALUE;
            flag = 0;
   
            // iterating over all the list
            for (i = 0; i < k; i++) {
                // if every element of list[i] is traversed then break the loop
                if (ptr[i] == n) {
                    flag = 1;
                    break;
                }
                // find minimum value among all the list elements pointing by the ptr[] array
                if (ptr[i] < n && arr[i][ptr[i]] < minval) {
                    minind = i; // update the index of the list
                    minval = arr[i][ptr[i]];
                }
                // find maximum value among all the list elements pointing by the ptr[] array
                if (ptr[i] < n && arr[i][ptr[i]] > maxval) {
                    maxval = arr[i][ptr[i]];
                }
            }
   
            // if any list exhaust we will not get any better answer, so break the while loop
            if (flag == 1) {
                break;
            }
   
            ptr[minind]++;
   
            // updating the minrange
            if ((maxval - minval) < minrange) {
                minel = minval;
                maxel = maxval;
                minrange = maxel - minel;
            }
        }
        document.write("The smallest range is ["+minel+", "+maxel+"]<br>");
}
 
// Driver program to test above function
let arr = [
    [4, 7, 9, 12, 15],
    [0, 8, 10, 14, 20],
    [6, 12, 16, 30, 50]
    ]
let k = arr.length;
findSmallestRange(arr, N, k);
 
 
// This code is contributed by unknown2108
 

输出：

The smallest range is [6 8]

• 复杂度分析：
  • 时间复杂度： O(n * k ² )，在长度为k的数组中找到最大值和最小值所需的时间为O(k)，遍历所有长度为n的数组（在最坏的情况下），时间复杂度为 O(n*k)，那么总时间复杂度为 O(n*k ² )。
  • 空间复杂度： O(k)，需要一个长度为 k 的额外数组，所以空间复杂度为 O(k)

高效的方法：方法保持不变，但可以通过使用 min-heap 或priority queue来降低时间复杂度。最小堆可用于在对数时间或 log k 时间而不是线性时间中找到最大值和最小值。其余方法保持不变。

• 算法：
  1. 创建一个 Min 堆来存储 k 个元素，每个数组一个元素，一个变量minrange初始化为最大值，并保留一个变量max来存储最大整数。
  2. 最初放置每个列表中每个元素的第一个元素，并将最大值存储在max 中。
  3. 重复以下步骤，直到至少一个列表耗尽：
     1. 要找到最小值或min ，请使用作为最小元素的 Min 堆的顶部或根。
     2. 如果当前 (max-min) 小于 minrange，现在更新 minrange。
     3. 从优先级队列中删除顶部或根元素，并从包含最小元素的列表中插入下一个元素，并使用插入的新元素更新最大值。
• 执行：

C++

// C++ program to finds out smallest range that includes
// elements from each of the given sorted lists.
#include 
using namespace std;
 
#define N 5
 
// A min heap node
struct MinHeapNode {
    // The element to be stored
    int element;
 
    // index of the list from which the element is taken
    int i;
 
    // index of the next element to be picked from list
    int j;
};
 
// Prototype of a utility function to swap two min heap nodes
void swap(MinHeapNode* x, MinHeapNode* y);
 
// A class for Min Heap
class MinHeap {
 
    // pointer to array of elements in heap
    MinHeapNode* harr;
 
    // size of min heap
    int heap_size;
 
public:
    // Constructor: creates a min heap of given size
    MinHeap(MinHeapNode a[], int size);
 
    // to heapify a subtree with root at given index
    void MinHeapify(int);
 
    // to get index of left child of node at index i
    int left(int i) { return (2 * i + 1); }
 
    // to get index of right child of node at index i
    int right(int i) { return (2 * i + 2); }
 
    // to get the root
    MinHeapNode getMin() { return harr[0]; }
 
    // to replace root with new node x and heapify() new root
    void replaceMin(MinHeapNode x)
    {
        harr[0] = x;
        MinHeapify(0);
    }
};
 
// Constructor: Builds a heap from a
// given array a[] of given size
MinHeap::MinHeap(MinHeapNode a[], int size)
{
    heap_size = size;
    harr = a; // store address of array
    int i = (heap_size - 1) / 2;
    while (i >= 0) {
        MinHeapify(i);
        i--;
    }
}
 
// A recursive method to heapify a subtree with root at
// given index. This method assumes that the subtrees
// are already heapified
void MinHeap::MinHeapify(int i)
{
    int l = left(i);
    int r = right(i);
    int smallest = i;
    if (l < heap_size && harr[l].element < harr[i].element)
        smallest = l;
    if (r < heap_size && harr[r].element < harr[smallest].element)
        smallest = r;
    if (smallest != i) {
        swap(harr[i], harr[smallest]);
        MinHeapify(smallest);
    }
}
 
// This function takes an k sorted lists in the form of
// 2D array as an argument. It finds out smallest range
// that includes elements from each of the k lists.
void findSmallestRange(int arr[][N], int k)
{
    // Create a min heap with k heap nodes. Every heap node
    // has first element of an list
    int range = INT_MAX;
    int min = INT_MAX, max = INT_MIN;
    int start, end;
 
    MinHeapNode* harr = new MinHeapNode[k];
    for (int i = 0; i < k; i++) {
        // Store the first element
        harr[i].element = arr[i][0];
 
        // index of list
        harr[i].i = i;
 
        // Index of next element to be stored
        // from list
        harr[i].j = 1;
 
        // store max element
        if (harr[i].element > max)
            max = harr[i].element;
    }
 
    // Create the heap
    MinHeap hp(harr, k);
 
    // Now one by one get the minimum element from min
    // heap and replace it with next element of its list
    while (1) {
        // Get the minimum element and store it in output
        MinHeapNode root = hp.getMin();
 
        // update min
        min = hp.getMin().element;
 
        // update range
        if (range > max - min + 1) {
            range = max - min + 1;
            start = min;
            end = max;
        }
 
        // Find the next element that will replace current
        // root of heap. The next element belongs to same
        // list as the current root.
        if (root.j < N) {
            root.element = arr[root.i][root.j];
            root.j += 1;
 
            // update max element
            if (root.element > max)
                max = root.element;
        }
 
        // break if we have reached end of any list
        else
            break;
 
        // Replace root with next element of list
        hp.replaceMin(root);
    }
 
    cout << "The smallest range is "
         << "["
         << start << " " << end << "]" << endl;
    ;
}
 
// Driver program to test above functions
int main()
{
    int arr[][N] = {
        { 4, 7, 9, 12, 15 },
        { 0, 8, 10, 14, 20 },
        { 6, 12, 16, 30, 50 }
    };
    int k = sizeof(arr) / sizeof(arr[0]);
 
    findSmallestRange(arr, k);
 
    return 0;
}

Java

// Java program to find out smallest
// range that includes elements from
// each of the given sorted lists.
class GFG {
 
    // A min heap node
    static class Node {
        // The element to be stored
        int ele;
 
        // index of the list from which
        // the element is taken
        int i;
 
        // index of the next element
        // to be picked from list
        int j;
 
        Node(int a, int b, int c)
        {
            this.ele = a;
            this.i = b;
            this.j = c;
        }
    }
 
    // A class for Min Heap
    static class MinHeap {
        Node[] harr; // array of elements in heap
        int size; // size of min heap
 
        // Constructor: creates a min heap
        // of given size
        MinHeap(Node[] arr, int size)
        {
            this.harr = arr;
            this.size = size;
            int i = (size - 1) / 2;
            while (i >= 0) {
                MinHeapify(i);
                i--;
            }
        }
 
        // to get index of left child
        // of node at index i
        int left(int i)
        {
            return 2 * i + 1;
        }
 
        // to get index of right child
        // of node at index i
        int right(int i)
        {
            return 2 * i + 2;
        }
 
        // to heapify a subtree with
        // root at given index
        void MinHeapify(int i)
        {
            int l = left(i);
            int r = right(i);
            int small = i;
            if (l < size && harr[l].ele < harr[i].ele)
                small = l;
            if (r < size && harr[r].ele < harr[small].ele)
                small = r;
            if (small != i) {
                swap(small, i);
                MinHeapify(small);
            }
        }
 
        void swap(int i, int j)
        {
            Node temp = harr[i];
            harr[i] = harr[j];
            harr[j] = temp;
        }
 
        // to get the root
        Node getMin()
        {
            return harr[0];
        }
 
        // to replace root with new node x
        // and heapify() new root
        void replaceMin(Node x)
        {
            harr[0] = x;
            MinHeapify(0);
        }
    }
 
    // This function takes an k sorted lists
    // in the form of 2D array as an argument.
    // It finds out smallest range that includes
    // elements from each of the k lists.
    static void findSmallestRange(int[][] arr, int k)
    {
        int range = Integer.MAX_VALUE;
        int min = Integer.MAX_VALUE;
        int max = Integer.MIN_VALUE;
        int start = -1, end = -1;
 
        int n = arr[0].length;
 
        // Create a min heap with k heap nodes.
        // Every heap node has first element of an list
        Node[] arr1 = new Node[k];
        for (int i = 0; i < k; i++) {
            Node node = new Node(arr[i][0], i, 1);
            arr1[i] = node;
 
            // store max element
            max = Math.max(max, node.ele);
        }
 
        // Create the heap
        MinHeap mh = new MinHeap(arr1, k);
 
        // Now one by one get the minimum element
        // from min heap and replace it with
        // next element of its list
        while (true) {
            // Get the minimum element and
            // store it in output
            Node root = mh.getMin();
 
            // update min
            min = root.ele;
 
            // update range
            if (range > max - min + 1) {
                range = max - min + 1;
                start = min;
                end = max;
            }
 
            // Find the next element that will
            // replace current root of heap.
            // The next element belongs to same
            // list as the current root.
            if (root.j < n) {
                root.ele = arr[root.i][root.j];
                root.j++;
 
                // update max element
                if (root.ele > max)
                    max = root.ele;
            }
            // break if we have reached
            // end of any list
            else
                break;
 
            // Replace root with next element of list
            mh.replaceMin(root);
        }
        System.out.print("The smallest range is [" + start + " " + end + "]");
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr[][] = { { 4, 7, 9, 12, 15 },
                        { 0, 8, 10, 14, 20 },
                        { 6, 12, 16, 30, 50 } };
 
        int k = arr.length;
 
        findSmallestRange(arr, k);
    }
}
 
// This code is contributed by nobody_cares

C＃

// C# program to find out smallest
// range that includes elements from
// each of the given sorted lists.
using System;
using System.Collections.Generic;
 
class GFG {
 
    // A min heap node
    public class Node {
        // The element to be stored
        public int ele;
 
        // index of the list from which
        // the element is taken
        public int i;
 
        // index of the next element
        // to be picked from list
        public int j;
 
        public Node(int a, int b, int c)
        {
            this.ele = a;
            this.i = b;
            this.j = c;
        }
    }
 
    // A class for Min Heap
    public class MinHeap {
        // array of elements in heap
        public Node[] harr;
 
        // size of min heap
        public int size;
 
        // Constructor: creates a min heap
        // of given size
        public MinHeap(Node[] arr, int size)
        {
            this.harr = arr;
            this.size = size;
            int i = (size - 1) / 2;
            while (i >= 0) {
                MinHeapify(i);
                i--;
            }
        }
 
        // to get index of left child
        // of node at index i
        int left(int i)
        {
            return 2 * i + 1;
        }
 
        // to get index of right child
        // of node at index i
        int right(int i)
        {
            return 2 * i + 2;
        }
 
        // to heapify a subtree with
        // root at given index
        void MinHeapify(int i)
        {
            int l = left(i);
            int r = right(i);
            int small = i;
            if (l < size && harr[l].ele < harr[i].ele)
                small = l;
            if (r < size && harr[r].ele < harr[small].ele)
                small = r;
            if (small != i) {
                swap(small, i);
                MinHeapify(small);
            }
        }
 
        void swap(int i, int j)
        {
            Node temp = harr[i];
            harr[i] = harr[j];
            harr[j] = temp;
        }
 
        // to get the root
        public Node getMin()
        {
            return harr[0];
        }
 
        // to replace root with new node x
        // and heapify() new root
        public void replaceMin(Node x)
        {
            harr[0] = x;
            MinHeapify(0);
        }
    }
 
    // This function takes an k sorted lists
    // in the form of 2D array as an argument.
    // It finds out smallest range that includes
    // elements from each of the k lists.
    static void findSmallestRange(int[, ] arr, int k)
    {
        int range = int.MaxValue;
        int min = int.MaxValue;
        int max = int.MinValue;
        int start = -1, end = -1;
 
        int n = arr.GetLength(0);
 
        // Create a min heap with k heap nodes.
        // Every heap node has first element of an list
        Node[] arr1 = new Node[k];
        for (int i = 0; i < k; i++) {
            Node node = new Node(arr[i, 0], i, 1);
            arr1[i] = node;
 
            // store max element
            max = Math.Max(max, node.ele);
        }
 
        // Create the heap
        MinHeap mh = new MinHeap(arr1, k);
 
        // Now one by one get the minimum element
        // from min heap and replace it with
        // next element of its list
        while (true) {
            // Get the minimum element and
            // store it in output
            Node root = mh.getMin();
 
            // update min
            min = root.ele;
 
            // update range
            if (range > max - min + 1) {
                range = max - min + 1;
                start = min;
                end = max;
            }
 
            // Find the next element that will
            // replace current root of heap.
            // The next element belongs to same
            // list as the current root.
            if (root.j < n) {
                root.ele = arr[root.i, root.j];
                root.j++;
 
                // update max element
                if (root.ele > max)
                    max = root.ele;
            }
            else
                break; // break if we have reached
            // end of any list
 
            // Replace root with next element of list
            mh.replaceMin(root);
        }
        Console.Write("The smallest range is [" + start + " " + end + "]");
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int[, ] arr = { { 4, 7, 9, 12, 15 },
                        { 0, 8, 10, 14, 20 },
                        { 6, 12, 16, 30, 50 } };
 
        int k = arr.GetLength(0);
 
        findSmallestRange(arr, k);
    }
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// Javascript program to find out smallest
// range that includes elements from
// each of the given sorted lists.
class Node
{
    constructor(a, b, c)
    {
        this.ele = a;
        this.i = b;
        this.j = c;
    }
}
 
// A class for Min Heap
class MinHeap
{
     
    // Array of elements in heap
    harr;
     
    // Size of min heap
    size;
     
    // Constructor: creates a min heap
    // of given size
    constructor(arr,size)
    {
        this.harr = arr;
        this.size = size;
        let i = Math.floor((size - 1) / 2);
         
        while (i >= 0)
        {
            this.MinHeapify(i);
            i--;
        }
    }
     
    // To get index of left child
    // of node at index i
    left(i)
    {
        return 2 * i + 1;
    }
     
    // To get index of right child
    // of node at index i
    right(i)
    {
        return 2 * i + 2;
    }
     
    // To heapify a subtree with
    // root at given index
    MinHeapify(i)
    {
        let l = this.left(i);
        let r = this.right(i);
        let small = i;
         
        if (l < this.size &&
                this.harr[l].ele <
                this.harr[i].ele)
            small = l;
        if (r < this.size &&
                this.harr[r].ele <
                this.harr[small].ele)
            small = r;
        if (small != i)
        {
            this.swap(small, i);
            this.MinHeapify(small);
        }
    }
      
    swap(i, j)
    {
         
        let temp = this.harr[i];
        this.harr[i] = this.harr[j];
        this.harr[j] = temp;
    }
     
    // To get the root
    getMin()
    {
        return this.harr[0];
    }
     
    // To replace root with new node x
    // and heapify() new root
    replaceMin(x)
    {
        this.harr[0] = x;
        this.MinHeapify(0);
    }
 
}
 
 
// This function takes an k sorted lists
// in the form of 2D array as an argument.
// It finds out smallest range that includes
// elements from each of the k lists.
function findSmallestRange(arr, k)
{
    let range = Number.MAX_VALUE;
    let min = Number.MAX_VALUE;
    let max = Number.MIN_VALUE;
    let start = -1, end = -1;
    let n = arr[0].length;
 
    // Create a min heap with k heap nodes.
    // Every heap node has first element of an list
    let arr1 = new Array(k);
    for(let i = 0; i < k; i++)
    {
        let node = new Node(arr[i][0], i, 1);
        arr1[i] = node;
 
        // Store max element
        max = Math.max(max, node.ele);
    }
 
    // Create the heap
    let mh = new MinHeap(arr1, k);
 
    // Now one by one get the minimum element
    // from min heap and replace it with
    // next element of its list
    while (true)
    {
         
        // Get the minimum element and
        // store it in output
        let root = mh.getMin();
 
        // Update min
        min = root.ele;
 
        // Update range
        if (range > max - min + 1)
        {
            range = max - min + 1;
            start = min;
            end = max;
        }
 
        // Find the next element that will
        // replace current root of heap.
        // The next element belongs to same
        // list as the current root.
        if (root.j < n)
        {
            root.ele = arr[root.i][root.j];
            root.j++;
 
            // Update max element
            if (root.ele > max)
                max = root.ele;
        }
         
        // Break if we have reached
        // end of any list
        else
            break;
 
        // Replace root with next element of list
        mh.replaceMin(root);
    }
    document.write("The smallest range is [" +
                   start + " " + end + "]");
}
 
// Driver Code
let arr = [ [ 4, 7, 9, 12, 15 ],
            [ 0, 8, 10, 14, 20 ],
            [ 6, 12, 16, 30, 50 ] ];
let k = arr.length;
 
findSmallestRange(arr, k);
 
// This code is contributed by rag2127
 

输出：

The smallest range is [6 8]

• 复杂度分析：
  • 时间复杂度： O(n * k *log k)。
    要在长度为 k 的最小堆中找到最大值和最小值所需的时间为 O(log k)，并且要遍历所有长度为 n 的 k 数组（在最坏的情况下），时间复杂度为 O(n*k)，那么总时间复杂度是 O(n * k *log k)。
  • 空间复杂度： O(k)。
    优先级队列将存储 k 个元素，因此空间复杂度为 O(k)

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