【分治】求逆序对个数并打印逆序对

如果用最hick的方法去求那么就是O(n^2)的复杂度，如果想优化的话，用归并排序的方法分治处理。

主要思想：总逆序=左边逆序+右边逆序+左边右边分别排序后的逆序 

#include<stdio.h>#include<stdlib.h>using namespace std;//用归并排序的思想来求，归并排序为O（nlogn）的时间复杂度，比暴力的O(n^2)会好。 int merge(int *a, int low, int mid, int high){    int l_size = mid-low+1;    int r_size = high-mid;    int *l_list = (int *)malloc(l_size*sizeof(int));    int *r_list = (int *)malloc(r_size*sizeof(int));    int i, j, k;    for(i=0; i<l_size; i++)        l_list[i] = a[low+i];    for(i=mid+1; i<=high; i++)        r_list[i-mid-1] = a[i];    k = low;    int count = 0;    for(i=0, j=0; i<l_size && j<r_size;)    {        if(l_list[i]<=r_list[j])            a[k++] = l_list[i++];        else        {           for(int tmp=i;tmp<l_size;tmp++)            printf("(%d,%d) ", l_list[tmp],r_list[j]); //打印逆序对;             a[k++] = r_list[j++];            count += l_size-i;  //从i到l_size-1，所以是l_size-1-i+1 = l_size-i个逆序对                  }    }    while(i<l_size)        a[k++] = l_list[i++];    while(j<r_size)        a[k++] = r_list[j++];    free(l_list);    free(r_list);    return count;}int reverse_order(int *a, int low, int high){    int count = 0;    if(low < high)    {        int mid = (low+high)>>1;        //总逆序=左边逆序+右边逆序+左边右边分别排序后的逆序         count += reverse_order(a, low, mid);        count += reverse_order(a, mid+1, high);        count += merge(a, low, mid, high);    }    return count;}int main(void){    //int a[] = {5,4,3,2,1};    int a[] = {1,2,4,3,0};    int n = sizeof(a)/sizeof(a[0]);        printf("原数组为 ：\n");    for(int i=0; i<n; i++)             printf("%d ",a[i]);    printf("\n打印逆序: \n");        int count=reverse_order(a, 0, n-1);     printf("\nreverse_order count: %d\n", count);         printf("排序后数组为 ：\n");    for(int i=0; i<n; i++)             printf("%d ", i,a[i]);    printf("\n");            system("pause");    return 0;}