[LintCode] Reverse Pairs 翻转对

 

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Count of Smaller Numbers After Self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Count of Smaller Numbers After Self，参见代码如下；

 

class Solution {
public:
    long long reversePairs(vector<int>& A) {
        long long res = 0;
        vector<int> 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;        
   }
};