思路: bit + 离散化. 实现: #include <bits/stdc++.h> using namespace std; class Solution { public: int sum(vector<int> & bit, int i) { ; while (i) { ans += bit[i]; i -= i & -i; } return ans; } void add(vector<int> & bit, int i, int x)…
315. Count of Smaller Numbers After Self class Solution { public: vector<int> countSmaller(vector<int>& nums) { int n = nums.size(); vector<int> v(n); for (int i = n - 1; i >= 0; --i) { int val = nums[i]; int L = i + 1, R = n - 1;…
You are given an integer array nums and you have to return a new countsarray. 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…
原题链接在这里:https://leetcode.com/problems/count-of-smaller-numbers-after-self/ 题目: 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 r…
原题链接在这里:https://leetcode.com/problems/count-of-smaller-numbers-after-self/ 题目: 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 r…
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: Given nums = [5, 2, 6, 1] To the right of 5 there are…
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: Given nums = [5, 2, 6, 1] To the right of 5 there are…
