《算法导论》6、计算一个数组中逆序数的个数（C++，使用合并排序改编）

来源：互联网发布：java里字段编辑：程序博客网时间：2024/05/23 11:15

#include <iostream>using namespace std;int merge(int A[], int p, int q, int r);int merge_sort(int A[], int p, int r);void main(){int a[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9 };cout << merge_sort(a, 0, 17) << endl;system("pause");}int merge(int A[], int p, int q, int r) {int i, j, k, inversions = 0;int n1 = q - p + 1;int n2 = r - q;int *L = new int[n1];int *R = new int[n2];for (i = 0; i < n1; i++) L[i] = A[p + i];for (j = 0; j < n2; j++) R[j] = A[q + j + 1];for (i = 0, j = 0, k = p; k <= r; k++) {if (i == n1) {A[k] = R[j++];}else if (j == n2) {A[k] = L[i++];}else if (L[i] <= R[j]) {A[k] = L[i++];}else {A[k] = R[j++];inversions += n1 - i;  //L比R小，计做一次逆序//这里巧妙地利用了一个前提条件，LR都是有序的}}return inversions;}int merge_sort(int A[], int p, int r) {if (p < r) {int inversions = 0;int q = (p + r) / 2;inversions += merge_sort(A, p, q);inversions += merge_sort(A, q + 1, r);inversions += merge(A, p, q, r);return inversions;}else {return 0;}}

0 0