【LeetCode 315】 Count of Smaller Numbers After Self

題目描述

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.

思路

從後往前遍歷，數字插入到新的數組中，二分查找當前數字的位置，即爲後邊有多少個小於當前數字的數。

代碼

class Solution {
public:
    vector<int> countSmaller(vector<int>& nums) {
        if (nums.empty()) return {};
        int n = nums.size();
        vector<int> res(n, 0);
        vector<int> sorted;
        
        for (int i=n-1; i>=0; --i) {
            auto it = lower_bound(sorted.begin(), sorted.end(), nums[i]);
            res[i] = it - sorted.begin();
            sorted.insert(it, nums[i]);
        }
        return res;
    }
};