For an array A, if i < j, and A [i] > A [j], called (A [i], A [j]) is a reverse pair.

return total of reverse pairs in A.

Example

Given A = [2, 4, 1, 3, 5] , (2, 1), (4, 1), (4, 3) are reverse pairs. return 3

 

这道题跟LeetCode上的那道是一样的，唯一的一点点的小区别是那道题是返回一个向量，表示出原数组中每一个数字的右边比其小的数的个数，而这道题让我们求翻转对的总数，其实就是把每个数字右边比其小的数的个数都加起来即可，具体讲解请参加之前那篇博客，参见代码如下；

 

class Solution {public:    long long reversePairs(vector
     
      & A) {        long long res = 0;        vector
      
        v;        for (int i = A.size() - 1; i >= 0; --i) {            int left = 0, right = v.size();            while (left < right) {                int mid = left + (right - left) / 2;                if (A[i] > v[mid]) left = mid + 1;                else right = mid;            }            v.insert(v.begin() + right, A[i]);            res += right;        }        return res;           }};