求數組逆序對

標籤：io   for   ar   cti   演算法   amp   new   c   

思路：類似歸併排序演算法，在合并已經有序的相鄰子數組的時候，計算前面數組相對於後面數組的逆序對數，整個遞迴過程可以算出所有逆序對
#include <stdio.h>void merge(int A[], int front, int middle, int end, int &count){    if (front >= end) return;    int i = front;    int j = middle + 1;    int k = 0;    int *p = new int[end - front + 1];    while (i <= middle && j <= end) {        if (A[i] < A[j]) {            p[k++] = A[i++];        } else {            count += middle - i + 1;            p[k++] = A[j++];        }    }    if (j <= end) while (j <= end) {        p[k++] = A[j++];        count += middle - i + 1;    }    if (i <= middle) while (i <= middle) p[k++] = A[i++];    printf("front:%d end:%d :", front, end);    for (int i = 0; i < k; ++i) {        printf("%d ", p[i]);        A[front + i] = p[i];    }    printf("\n");    delete [] p;}void merge_sort(int A[], int front, int end, int &count){    if (front >= end) return;    int middle = (front + end) / 2;    merge_sort(A, front, middle, count);    merge_sort(A, middle + 1, end, count);    merge(A, front, middle, end, count);}int main(){    int count = 0;    int rand[] = {3,1,5,23,2,32,56,76,65,34,2,2,333,1,0};    merge_sort(rand, 0, 4,  count);    for (int i = 0; i < 5; ++i) {        printf("%d ", rand[i]);    }    printf("\n");    printf("%d\n", count);    return 0;}