You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i]
is the number of smaller elements to the right of nums[i]
.
Example:
Input: [5,2,6,1]
Output: [2,1,1,0]
Explanation:
To the right of 5 there are 2 smaller elements (2 and 1).
To the right of 2 there is only 1 smaller element (1).
To the right of 6 there is 1 smaller element (1).
To the right of 1 there is 0 smaller element.
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基本思路 从后向前遍历,每遍历一个数把他放到一个数组里面,有序插入, 用二分是最快的.
用stl的函数用节省不少代码
class Solution { public: vector<int> countSmaller(vector<int>& nums) { vector<int> after,res(nums.size()); for(int i=nums.size()-1;i>=0;--i) { auto iter=lower_bound(after.begin(),after.end(),nums[i]); res[i]=iter-after.begin(); after.insert(iter,nums[i]); } return res; } };