归并排序

/*
 * 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。