统计序列中逆序对数量C语言

  <<算法导论>>里面的一道题.今天上午看的东西.

  逆序对,举个例子,A[0] > A[1],则A[0]和A[1]是一对逆序对.呵呵.按照书上所说,用归并排序实现.我给实现了.

  思路很清晰,很享受研究算法的过程.

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/* 2-4-11-04-06-13.16.c -- 第二章第四题 */ #include <stdio.h> #include <stdlib.h> #define SIZE (27) int main (void) ; int mergeSort (int * const array, const int left, const int right) ; int merge (int * const array, const int left, const int middle, const int right) ; void printResult (const int * const array, const int size) ; int main (void) { int array[SIZE] = {4, 2, 9, 7, 6, 3, 1, 5, 7, 54, 12, 87, 14, 62, 15, 0, 54, 33, 41, 87, 56, 43, 76, 53, 21, 46, 88} ; int size = SIZE ; printf ("%d/n", mergeSort (array, 0, size - 1)) ; printResult (array, size) ; return 0 ; } int mergeSort (int * const array, const int left, const int right) { static int count = 0 ; int middle ; if (left < right) { middle = (left + right) / 2 ; mergeSort (array, left, middle) ; mergeSort (array, middle + 1, right) ; count += merge (array, left, middle, right) ; return count ; } else return 0 ; } int merge (int * const array, const int left, const int middle, const int right) { int * first, * second ; int fLenth, sLenth ; int i, j, k ; int count ; fLenth = middle - left + 1 ; sLenth = right - middle ; first = (int *) malloc (sizeof (int) * fLenth) ; second = (int *) malloc (sizeof (int) * sLenth) ; for (i = 0; i < fLenth; i++) first[i] = array[left + i] ; for (j = 0; j < sLenth; j++) second[j] = array[middle + j + 1] ; count = 0 ; i = j = 0 ; for (k = left; k <= right; k++) { if (first[i] <= second[j]) array[k] = first[i++] ; else { array[k] = second[j++] ; count += fLenth - i ; } if (i == fLenth) { k++ ; while (k <= right) array[k++] = second[j++] ; break ; } else if (j == sLenth) { k++ ; while (k <= right) array[k++] = first[i++] ; break ; } } free (first) ; free (second) ; return count ; } void printResult (const int * const array, const int size) { int i ; for (i = 0; i < size; i++) printf ("%-3d", array[i]) ; putchar ('/n') ; }