/*题目：在数组中的两个数字如果见面一个数字大于后面的数字，则这两个数字组成一个逆序对。输入一个数组，求这个数组中的逆序对的总数*/#include <iostream>using namespace std;int ReversePairsCore(int data[], int copy[], int start, int end){ //递归出口 if(start == end) { copy[start] = data[start]; return 0; } int length = (end - start) / 2; //递归调用，并利用归并排序实现排序结排序好的结果放在data[]数组中，所以copy和data位置调换 //(递归时要用到copy数组作为中转将其转存到data中)。 //利用递归将数组分为两部分 int left = ReversePairsCore(copy, data, start, start + length); int right = ReversePairsCore(copy, data, start + length + 1, end); //归并排序实现 int i = start + length; int j = end; int indexCopy = end; int count = 0; while(i >= start && j >= start + length + 1) { if(data[i] > data[j]) { copy[indexCopy--] = data[i--]; count += j - start - length; } else { copy[indexCopy--] = data[j--]; } } for(; i >= start; --i) copy[indexCopy--] = data[i]; for(; j >= start + length + 1; --j) copy[indexCopy--] = data[j]; return left + right + count;}//int ReversePairs(int data[], int length){ if(data == NULL || length < 0) return 0; int *copy = new int[length]; //初始化辅助数组为和data数组相同 for(int i = 0; i < length; ++i) copy[i] = data[i]; int count = ReversePairsCore(data, copy, 0, length - 1); delete [] copy; return count;}//================测试代码=================void Test(int data[], int length, int expected){ if(ReversePairs(data, length) == expected) cout << "Passed!" << endl; else cout << "Failed!" << endl;}//=============================================int main(){ //递增数组 int data1[] = {1, 2, 3, 4, 5, 6}; int expected1 = 0; Test(data1, sizeof(data1) / sizeof(int), expected1); //递减数组 int data2[] = {6, 5, 4, 3, 2, 1}; int expected2 = 15; Test(data2, sizeof(data2) / sizeof(int), expected2); //未经排序的数组 int data3[] = {1, 2, 3, 4, 7, 6, 5}; int expected3 = 3; Test(data3, sizeof(data3) / sizeof(int), expected3); //数组中包含重复的数字 int data4[] = {1, 2, 1, 2, 1}; int expected4 = 3; Test(data4, sizeof(data4) / sizeof(int), expected4); //数组中只有两个数字，递增排序 int data5[] ={1, 2}; int expected5 = 0; Test(data5, sizeof(data5) / sizeof(int), expected5); //数组中只有两个数字， 递减排序 int data6[] = {2, 1}; int expected6 = 1; Test(data6, sizeof(data6) / sizeof(int), expected6); //数组中只有一个数字 int data7[] = {1}; int expected7 = 0; Test(data7, sizeof(data7) / sizeof(int), expected7); //输入空指针 Test(NULL, 0, 0); return 0;}
