k路归并 O(nlogk)

题目

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处理k-路合并操作问题，并分析时间复杂度。

算法思想

采用分治法归并排序，归并两个有序数组时间复杂度为O(n)，将K个有序数组分治归并时间复杂度为O(logk)，算法整体时间复杂度为O(nlogk)，程序里用到了vector向量容器。

#include <iostream>#include <vector>using namespace std;vector<int> mergeTowArrays(vector<int>A,vector<int>B){    vector<int>temp;    temp.resize(A.size() + B.size());    int index = 0, j = 0, i = 0;    while (i < A.size() && j < B.size())    {        if (A[i] < B[j])            temp[index++] = A[i++];        else            temp[index++] = B[j++];    }        while (i < A.size())            temp[index++] = A[i++];        while (j < B.size())            temp[index++] = B[j++];        return temp;}vector<int> kMergeSort(vector<vector<int>>A, int start, int end){    if (start >= end)        return A[start];    int mid = start + (end - start) / 2;    vector<int>Left = kMergeSort(A, start, mid);    vector<int>Right = kMergeSort(A, mid + 1, end);    return mergeTowArrays(Left, Right);}vector<int> mergeSortArrays(vector <vector<int>>A){    vector<int>temp;    if (A.empty() || A.size() == 0 || A[0].size() == 0)        return temp;    temp = kMergeSort(A, 0, A.size() - 1);    return temp;}int main(void){    int k,n;    cin >> k >> n;    vector<vector<int>>A(k);    for (int i = 0; i < k; i++)    {        A[i].resize(n);    }    for (int i = 0; i < A.size(); i++)    {        for (int j = 0; j < A[0].size(); j++)            cin >> A[i][j];    }    vector<int>result;    result = mergeSortArrays(A);    for (int i = 0; i < result.size(); i++)    {        cout << result[i] << " ";    }    cout << endl;    system("pause");    return 0;}