假定有k个有序数组，每个数组中含有n个元素，您的任务是将它们合并为单独的一个有序数组，该数组共有kn个元素。
设计和实现 一个有效的分治算法解决k-路合并操作问题，并分析时间复杂度。

思路

分治算法即将有序数组两两归并
相关归并代码源自网上博主写的java代码，java的数组可以直接求length，但C++的数组要手动传入length很麻烦……故改写成C++费了一些时间

#include <iostream>
#define N 5
#define INDEX_COUNT 5
using namespace std;
int INDEX[INDEX_COUNT][N] =
                            { 1,4,5,8,9,
                            2,6,8,9,10,
                            3,5,5,7,8,
                            4,5,6,7,12,
                            1,5,8,9,10 };

int MERGED_INDEXS[N*INDEX_COUNT];

void showArray(int a[], int length)
{
    for (int i = 0; i < length; i++)
    {
        cout << a[i] << " ";
    }
    cout << "\n";
}


int* mergeTwoArrays(int* A, int* B,int lengthA,int lengthB) {
    int* temp = new int[lengthA + lengthB];
    int index = 0, i = 0, j = 0;
    while (i < lengthA && j < lengthB) {
        if (A[i] < B[j]) {
            temp[index++] = A[i++];
        }
        else {
            temp[index++] = B[j++];
        }
    }
    while (i < lengthA) {
        temp[index++] = A[i++];
    }
    while (j < lengthB) {
        temp[index++] = B[j++];
    }
    return temp;
}

// 分治递归深度为O(log k), 每层合并时间复杂度O(n)
int* kMergeSort(int arrays[INDEX_COUNT][N], int start, int end) {
    if (start >= end) {
        return arrays[start];
    }
    int mid = start + (end - start) / 2;
    int* left = kMergeSort(arrays, start, mid);
    int* right = kMergeSort(arrays, mid + 1, end);
    cout << "归并INDEX" << start << "和INDEX" << end<<endl;
    return mergeTwoArrays(left, right, N*(mid - start + 1), N*(end - mid));
}

int* mergekSortedArrays(int arrays[INDEX_COUNT][N],int length) {
    int* list = new int[N*INDEX_COUNT];
    if (arrays == NULL) {
        return list;
    }
    int* ans = kMergeSort(arrays, 0, length - 1);
    showArray(ans, N*INDEX_COUNT);
    return list;
}




int main(void) {
    for (int i = 0; i < INDEX_COUNT; i++) {
        cout << "数组" << i << ":";
        showArray(INDEX[i], N);
    }
    mergekSortedArrays(INDEX, INDEX_COUNT);
    //showArray(mergeTwoArrays(INDEX[0], INDEX[1], N, N), N + N);
    system("pause");
    return 0;
}