剑指offer:数组中的逆序对class Solution {public:void solve(int x, int y, vector<int> &A, vector<int> &B, int &ans){//B为临时空间if(y - x > 1){int mid = x + (y - x)/2, left = x, right = mid, i = x;solve(x, mid, A, B, ans);solve(mid, y, A, B, ans);int zuo = mid - x;while(left < mid || right < y){if(right >= y || (left < mid && A[left] < A[right])){B[i++] = A[left++];zuo--;}else {B[i++] = A[right++];ans = (ans+zuo)00000007;}}}for(int i = x; i < y; i++) A[i] = B[i];}int InversePairs(vector<int> data) {vector<int> tmp = data;int ans = 0;solve(0, data.size(), data, tmp, ans);return ans;}};