给定 k 个可能不同大小的排序数组，合并它们并打印排序后的输出。
例子：
Input: k = 3
arr[][] = { {1, 3},
{2, 4, 6},
{0, 9, 10, 11}} ;
Output: 0 1 2 3 4 6 9 10 11
Input: k = 2
arr[][] = { {1, 3, 20},
{2, 4, 6}} ;
Output: 1 2 3 4 6 20 我们在 Merge k sorted arrays | 中讨论了一个适用于所有相同大小数组的解决方案。设置 1。
一个简单的解决方案是创建一个输出数组，然后将所有数组一一复制到其中。最后，使用排序输出数组。这种方法需要 O(N Logn N) 时间，其中 N 是所有元素的计数。
一个有效的解决方案是使用堆数据结构。基于堆的解决方案的时间复杂度是 O(N Log k)。
1. 创建一个输出数组。
2. 创建一个大小为 k 的最小堆并将所有数组中的第一个元素插入堆中
3. 优先级队列不为空时重复以下步骤。
…..a) 从堆中移除最小元素（最小值总是在根）并将其存储在输出数组中。
…..b) 从提取元素的数组中插入下一个元素。如果数组没有更多元素，则什么都不做。
C++
// C++ program to merge k sorted arrays
// of size n each.
#include
using namespace std;
// A pair of pairs, first element is going to
// store value, second element index of array
// and third element index in the array.
typedef pair > ppi;
// This function takes an array of arrays as an
// argument and all arrays are assumed to be
// sorted. It merges them together and prints
// the final sorted output.
vector mergeKArrays(vector > arr)
{
vector output;
// Create a min heap with k heap nodes. Every
// heap node has first element of an array
priority_queue, greater > pq;
for (int i = 0; i < arr.size(); i++)
pq.push({ arr[i][0], { i, 0 } });
// Now one by one get the minimum element
// from min heap and replace it with next
// element of its array
while (pq.empty() == false) {
ppi curr = pq.top();
pq.pop();
// i ==> Array Number
// j ==> Index in the array number
int i = curr.second.first;
int j = curr.second.second;
output.push_back(curr.first);
// The next element belongs to same array as
// current.
if (j + 1 < arr[i].size())
pq.push({ arr[i][j + 1], { i, j + 1 } });
}
return output;
}
// Driver program to test above functions
int main()
{
// Change n at the top to change number
// of elements in an array
vector > arr{ { 2, 6, 12 },
{ 1, 9 },
{ 23, 34, 90, 2000 } };
vector output = mergeKArrays(arr);
cout << "Merged array is " << endl;
for (auto x : output)
cout << x << " ";
return 0;
} Python
# merge function merge two arrays
# of different or same length
# if n = max(n1, n2)
# time complexity of merge is (o(n log(n)))
from heapq import merge
# function for meging k arrays
def mergeK(arr, k):
l = arr[0]
for i in range(k-1):
# when k = 0 it merge arr[1]
# with arr[0] here in l arr[0]
# is stored
l = list(merge(l, arr[i + 1]))
return l
# for printing array
def printArray(arr):
print(*arr)
# driver code
arr =[[2, 6, 12 ],
[ 1, 9 ],
[23, 34, 90, 2000 ]]
k = 3
l = mergeK(arr, k)
printArray(l)C#
// C# program to merge k sorted arrays
// of size n each.
using System;
using System.Collections.Generic;
class GFG
{
// This function takes an array of arrays as an
// argument and all arrays are assumed to be
// sorted. It merges them together and prints
// the final sorted output.
static List mergeKArrays(List> arr)
{
List output = new List();
// Create a min heap with k heap nodes. Every
// heap node has first element of an array
List>> pq =
new List>>();
for (int i = 0; i < arr.Count; i++)
pq.Add(new Tuple>(arr[i][0],
new Tuple(i, 0)));
pq.Sort();
// Now one by one get the minimum element
// from min heap and replace it with next
// element of its array
while (pq.Count > 0) {
Tuple> curr = pq[0];
pq.RemoveAt(0);
// i ==> Array Number
// j ==> Index in the array number
int i = curr.Item2.Item1;
int j = curr.Item2.Item2;
output.Add(curr.Item1);
// The next element belongs to same array as
// current.
if (j + 1 < arr[i].Count)
{
pq.Add(new Tuple>(arr[i][j + 1],
new Tuple(i, j + 1)));
pq.Sort();
}
}
return output;
}
// Driver code
static void Main()
{
// Change n at the top to change number
// of elements in an array
List> arr = new List>();
arr.Add(new List(new int[]{2, 6, 12}));
arr.Add(new List(new int[]{1, 9}));
arr.Add(new List(new int[]{23, 34, 90, 2000}));
List output = mergeKArrays(arr);
Console.WriteLine("Merged array is ");
foreach(int x in output)
Console.Write(x + " ");
}
}
// This code is contributed by divyeshrabadiya07. 输出：
Merged array is
1 2 6 9 12 23 34 90 2000如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。