对接近排序（或 K 排序）的数组进行排序

给定一个包含 n 个元素的数组，其中每个元素与其目标位置的距离至多 k，设计一个算法，在 O(n log k) 时间内进行排序。例如，让我们考虑 k 是 2，排序数组中索引为 7 的元素可以位于给定数组中的索引 5、6、7、8、9。

例子：

Input : arr[] = {6, 5, 3, 2, 8, 10, 9}
            k = 3 
Output : arr[] = {2, 3, 5, 6, 8, 9, 10}

Input : arr[] = {10, 9, 8, 7, 4, 70, 60, 50}
         k = 4
Output : arr[] = {4, 7, 8, 9, 10, 50, 60, 70}

我们可以使用插入排序来有效地对元素进行排序。以下是标准插入排序的 C 代码。

C++

/* Function to sort an array using insertion sort*/
void insertionSort(int A[], int size)
{
   int i, key, j;
   for(i = 1; i < size; i++)
   {
       key = A[i];
       j = i - 1;
 
       /* Move elements of A[0..i-1], that are
          greater than key, to one
          position ahead of their current position.
          This loop will run at most k times */
       while (j >= 0 && A[j] > key)
       {
           A[j + 1] = A[j];
           j = j - 1;
       }
       A[j + 1] = key;
   }
}
 
// This code is contributed by Shivani

C

/* Function to sort an array using insertion sort*/
void insertionSort(int A[], int size)
{
   int i, key, j;
   for (i = 1; i < size; i++)
   {
       key = A[i];
       j = i-1;
 
       /* Move elements of A[0..i-1], that are
          greater than key, to one
          position ahead of their current position.
          This loop will run at most k times */
       while (j >= 0 && A[j] > key)
       {
           A[j+1] = A[j];
           j = j-1;
       }
       A[j+1] = key;
   }
}

Java

/* Function to sort an array using insertion sort*/
static void insertionSort(int A[], int size)
{
    int i, key, j;
    for (i = 1; i < size; i++)
    {
        key = A[i];
        j = i-1;
 
        /* Move elements of A[0..i-1], that
            are greater than key, to one
            position ahead of their current position.
            This loop will run at most k times */
        while (j >= 0 && A[j] > key)
        {
            A[j+1] = A[j];
            j = j-1;
        }
        A[j+1] = key;
    }
}

Python3

# Function to sort an array using
# insertion sort
 
def insertionSort(A, size):
    i, key, j = 0, 0, 0
    for i in range(size):
        key = A[i]
        j = i-1
 
        # Move elements of A[0..i-1], that are
        # greater than key, to one position
        # ahead of their current position.
        # This loop will run at most k times
        while j >= 0 and A[j] > key:
            A[j + 1] = A[j]
            j = j - 1
        A[j + 1] = key

C#

/* C# Function to sort an array using insertion sort*/
static void insertionSort(int A[], int size)
{
    int i, key, j;
   
    for (i = 1; i < size; i++) {
        key = A[i];
        j = i - 1;
 
        /* Move elements of A[0..i-1], that are greater than
           key, to one position ahead of their current
           position. This loop will run at most k times */
        while (j >= 0 && A[j] > key) {
            A[j + 1] = A[j];
            j = j - 1;
        }
        A[j + 1] = key;
    }
}

Javascript

/* Function to sort an array using insertion sort*/
function insertionSort(A, size)
{
   var i, key, j;
   for (i = 1; i < size; i++)
   {
       key = A[i];
       j = i-1;
 
       /* Move elements of A[0..i-1], that are
          greater than key, to one
          position ahead of their current position.
          This loop will run at most k times */
       while (j >= 0 && A[j] > key)
       {
           A[j+1] = A[j];
           j = j-1;
       }
       A[j+1] = key;
   }
}
 
// This code is contributed by itsok.

内循环最多运行 k 次。要将每个元素移动到正确的位置，最多需要移动 k 个元素。所以整体复杂度将是 O(nk)

借助堆数据结构，我们可以更有效地对此类数组进行排序。以下是使用 Heap 的详细过程。
1) 使用前 k+1 个元素创建大小为 k+1 的最小堆。这将花费 O(k) 时间（请参阅此 GFact）
2) 将堆中的最小元素一一取出，放入结果数组，并从剩余元素中添加一个新元素到堆中。
删除一个元素并将一个新元素添加到最小堆将花费 log k 时间。所以整体复杂度将是 O(k) + O((nk) * log(k))。

C++

// A STL based C++ program to sort a nearly sorted array.
#include 
using namespace std;
 
// Given an array of size n, where every element
// is k away from its target position, sorts the
// array in O(n logk) time.
int sortK(int arr[], int n, int k)
{
     
    // Insert first k+1 items in a priority queue (or min
    // heap)
    //(A O(k) operation). We assume, k < n.
  //if size of array = k i.e k away from its target position
  //then
    int size;
    size=(n==k)?k:k+1;
    priority_queue, greater > pq(arr, arr +size);
 
    // i is index for remaining elements in arr[] and index
    // is target index of for current minimum element in
    // Min Heap 'pq'.
    int index = 0;
    for (int i = k + 1; i < n; i++) {
        arr[index++] = pq.top();
        pq.pop();
        pq.push(arr[i]);
    }
 
    while (pq.empty() == false) {
        arr[index++] = pq.top();
        pq.pop();
    }
}
 
// A utility function to print array elements
void printArray(int arr[], int size)
{
    for (int i = 0; i < size; i++)
        cout << arr[i] << " ";
    cout << endl;
}
 
// Driver program to test above functions
int main()
{
    int k = 3;
    int arr[] = { 2, 6, 3, 12, 56, 8 };
    int n = sizeof(arr) / sizeof(arr[0]);
    sortK(arr, n, k);
 
    cout << "Following is sorted array" << endl;
    printArray(arr, n);
 
    return 0;
}

Java

// A java program to sort a nearly sorted array
import java.util.Iterator;
import java.util.PriorityQueue;
 
class GFG {
    private static void kSort(int[] arr, int n, int k)
    {
 
        // min heap
        PriorityQueue priorityQueue
            = new PriorityQueue<>();
 
        // add first k + 1 items to the min heap
        for (int i = 0; i < k + 1; i++) {
            priorityQueue.add(arr[i]);
        }
 
        int index = 0;
        for (int i = k + 1; i < n; i++) {
            arr[index++] = priorityQueue.peek();
            priorityQueue.poll();
            priorityQueue.add(arr[i]);
        }
 
        Iterator itr = priorityQueue.iterator();
 
        while (itr.hasNext()) {
            arr[index++] = priorityQueue.peek();
            priorityQueue.poll();
        }
    }
 
    // A utility function to print the array
    private static void printArray(int[] arr, int n)
    {
        for (int i = 0; i < n; i++)
            System.out.print(arr[i] + " ");
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int k = 3;
        int arr[] = { 2, 6, 3, 12, 56, 8 };
        int n = arr.length;
        kSort(arr, n, k);
        System.out.println("Following is sorted array");
        printArray(arr, n);
    }
}
 
// This code is contributed by
// Manpreet Singh(manpreetsngh294)

蟒蛇3

# A Python3 program to sort a
# nearly sorted array.
 
from heapq import heappop, heappush, heapify
 
 
# A utility function to print
# array elements
def print_array(arr: list):
    for elem in arr:
        print(elem, end=' ')
 
# Given an array of size n, where every
# element is k away from its target
# position, sorts the array in O(nLogk) time.
 
 
def sort_k(arr: list, n: int, k: int):
    """
    :param arr: input array
    :param n: length of the array
    :param k: max distance, which every
     element is away from its target position.
    :return: None
    """
    # List of first k+1 items
    heap = arr[:k + 1]
 
    # using heapify to convert list
    # into heap(or min heap)
    heapify(heap)
 
    # "rem_elmnts_index" is index for remaining
    # elements in arr and "target_index" is
    # target index of for current minimum element
    # in Min Heap "heap".
    target_index = 0
    for rem_elmnts_index in range(k + 1, n):
        arr[target_index] = heappop(heap)
        heappush(heap, arr[rem_elmnts_index])
        target_index += 1
 
    while heap:
        arr[target_index] = heappop(heap)
        target_index += 1
 
 
# Driver Code
k = 3
arr = [2, 6, 3, 12, 56, 8]
n = len(arr)
sort_k(arr, n, k)
 
print('Following is sorted array')
print_array(arr)
 
# This code is contributed by
# Veerat Beri(viratberi)

C＃

// A C# program to sort a nearly sorted array
using System;
using System.Collections.Generic;
class GFG {
 
    static void kSort(int[] arr, int n, int k)
    {
 
        // min heap
        List priorityQueue = new List();
 
        // add first k + 1 items to the min heap
        for (int i = 0; i < k + 1; i++) {
            priorityQueue.Add(arr[i]);
        }
 
        priorityQueue.Sort();
 
        int index = 0;
        for (int i = k + 1; i < n; i++) {
            arr[index++] = priorityQueue[0];
            priorityQueue.RemoveAt(0);
            priorityQueue.Add(arr[i]);
            priorityQueue.Sort();
        }
 
        int queue_size = priorityQueue.Count;
 
        for (int i = 0; i < queue_size; i++) {
            arr[index++] = priorityQueue[0];
            priorityQueue.RemoveAt(0);
        }
    }
 
    // A utility function to print the array
    static void printArray(int[] arr, int n)
    {
        for (int i = 0; i < n; i++)
            Console.Write(arr[i] + " ");
    }
 
    // Driver code
    static void Main()
    {
        int k = 3;
        int[] arr = { 2, 6, 3, 12, 56, 8 };
        int n = arr.Length;
        kSort(arr, n, k);
        Console.WriteLine("Following is sorted array");
        printArray(arr, n);
    }
}
 
// This code is contributed by divyeshrabadiya07.

Javascript

// A javascript program to sort a nearly sorted array
 
function kSort(arr,n,k)
{
    let priorityQueue=[];
    // add first k + 1 items to the min heap
        for (let i = 0; i < k + 1; i++) {
            priorityQueue.push(arr[i]);
        }
        priorityQueue.sort(function(a,b){return a-b;});
  
        let index = 0;
        for (let i = k + 1; i < n; i++) {
            arr[index++] = priorityQueue[0];
            priorityQueue.shift();
            priorityQueue.push(arr[i]);
            priorityQueue.sort(function(a,b){return a-b;});
        }
  
         
  
        while (priorityQueue.length != 0) {
            arr[index++] = priorityQueue[0];
            priorityQueue.shift();
        }
}
 
// A utility function to print the array
function printArray(arr,n)
{
    for (let i = 0; i < n; i++)
            document.write(arr[i] + " ");
}
 
// Driver Code
let k = 3;
let arr = [ 2, 6, 3, 12, 56, 8 ];
let n = arr.length;
kSort(arr, n, k);
document.write("Following is sorted array<br>");
printArray(arr, n);
 
// This code is contributed by unknown2108

输出：

Following is sorted array
2 3 6 8 12 56

基于最小堆的方法需要 O(k) + O((m) * log(k)) 时间，其中 m = n – k 并使用 O(k) 辅助空间。

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