归并排序

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/* * name : merge sort * author : sangoly * O(nlgn) * S(n) * date : 2014/4/15 */#include <iostream>#include <cstdlib>#include <limits>using namespace std;void merge(int a[], int start, int middle, int end){          int max = numeric_limits<int>::max();     int tempArray1[middle-start+2];//include the start to middle     int tempArray2[end-middle+1];//include the middle+1 to end          for (int i=0; i<middle-start+1; i++)         tempArray1[i] = a[start+i];     tempArray1[middle-start+1] = max;//As the soldier          for (int i=0; i<end-middle; i++)         tempArray2[i] = a[middle+1+i];     tempArray2[end-middle] = max;//As the soldier          int index1 = 0;     int index2 = 0;          for (int k=start; k<=end; k++)         a[k] = tempArray1[index1] > tempArray2[index2] ? (tempArray2[index2++]) : (tempArray1[index1++]);}void merge_sort(int a[], int start, int end){     if (start >= end)        return;     int middle = (start+end)/2;     merge_sort(a, start, middle);     merge_sort(a, middle + 1, end);     merge(a, start, middle, end);}int main(int argc, const char **argv) {    int a[] = {3, 41, 52, 26, 38, 57, 9, 49};    merge_sort(a, 0, 7);    for (int i=0; i<8; i++)        cout<<a[i]<<" ";    cout<<endl;    system("pause");    return 0;}

归并排序，利用分治法的思想将一个大问题分解为形式等价的小问题的集合，递归求解。如果在插入到原数组的过程中采用插入排序的算法，也可以达到以常量组的空间复杂度完成此项任务，但这是以时间换空间的一种做法，时间性能上会有一定损耗。

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