1. 题目

给定一个数组 nums ，如果 i < j 且 nums[i] > 2*nums[j] 我们就将 (i, j) 称作一个重要翻转对。

你需要返回给定数组中的重要翻转对的数量。

输入: [1,3,2,3,1]
输出: 2输入: [2,4,3,5,1]
输出: 3

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/reverse-pairs
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2. 归并排序求解

• 参考归并排序

class Solution {int count = 0;int *temp;
public:int reversePairs(vector<int>& nums) {temp = new int[nums.size()];rp(nums,0,nums.size()-1);return count;}void rp(vector<int>& nums, int left, int right) {if(left >= right)return;int mid = (left+right)/2;rp(nums,left,mid);rp(nums,mid+1,right);merge(nums,left,mid,right);}void merge(vector<int>& nums, int left, int mid, int right){int i = left, j = mid+1, k;while(i <= mid && j <= right){if((long)nums[i] > 2*long(nums[j])){count += mid-i+1;j++;}elsei++;}//先处理问题//下面再排序i = left, j = mid+1, k = left;while(i <= mid && j <= right){if(nums[i] <= nums[j])temp[k++] = nums[i++];elsetemp[k++] = nums[j++];}while(i <= mid)temp[k++] = nums[i++];while(j <= right)temp[k++] = nums[j++];for(i = left; i <= right; ++i)nums[i] = temp[i];}
};