剑指offer:数组中的逆序对

题目描述

在数组中的两个数字，如果前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，求出这个数组中的逆序对的总数。

class Solution {public:    int InversePairs(vector<int> data) {        int len = data.size();        if (len == 0){            return 0;        }        vector<int> tmp(len);        int res = mergeSort(data, tmp, 0, len - 1);        return res;    }private:    int mergeSort(vector<int> &data, vector<int> &tmp, int left, int right){        if (left == right){            return 0;        }        int mid = left + (right - left) / 2;        int leftCnt = mergeSort(data, tmp, left, mid);        int rightCnt = mergeSort(data, tmp, mid + 1, right);        int i = mid;        int j = right;        int tmpIndex = right;        int cnt = 0;        while (i >= left && j >= mid + 1){            if (data[i] > data[j]){                tmp[tmpIndex--] = data[i--];                //难点是怎么算cnt                cnt += j - mid;            }            else{                tmp[tmpIndex--] = data[j--];            }        }        for (; i >= left; i--){            tmp[tmpIndex--] = data[i--];        }        for (; j >= mid + 1; j--){            tmp[tmpIndex--] = data[j--];        }        for (int i = left; i <= right; i++){            data[i] = tmp[i];        }        return leftCnt + rightCnt + cnt;    }};