You are given an integer array nums and you have to return a new counts array. The countsarray 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…
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…
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;…
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. Thecounts 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…
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:[5,2,6,1] Explanation: To the…
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 1: Input: nums = [5,2,6,1] Output: [2,1,1,0] Explanati…
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…
https://leetcode.com/problems/count-of-smaller-numbers-after-self/ 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…
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…
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)…
Given a binary tree containing digits from0-9only, each root-to-leaf path could represent a number. An example is the root-to-leaf path1->2->3which represents the number123. Find the total sum of all root-to-leaf numbers. For example, 1 / \ 2 3 The…
Given an integer array nums, return the number of range sums that lie in [lower, upper] inclusive.Range sum S(i, j) is defined as the sum of the elements in nums between indices i and j (i ≤ j), inclusive. Note:A naive algorithm of O(n2) is trivial.…
Almost identical to LintCode "Count of Smaller Number before Self". Corner case needs to be taken care of. class Solution { ////////////////// // Fenwick Tree // vector<long long> ft; void update(int i, long long x) { ) { ft[] ++; return;…
链接:http://leetcode.com/onlinejudge Add Two Numbers You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a lin…
原题 原题链接 Description: Count the number of prime numbers less than a non-negative number, n. 计算小于非负数n的素数个数. 思路 这题用埃拉托斯特尼筛法来做效果比较好,普通的方法基本会TLE.但是在用了埃拉托斯特尼筛法之后,还有一些细节值得注意: (1)首先我们该用 i*i<=n 替代 i<=sqrt(n) 来避免使用 sqrt() ,因为sqrt()的操作是比较expensive的. (2)当要表示两个状…
[题目描述] Give you an integer array (index from 0 to n-1, where n is the size of this array, data value from 0 to 10000) . For each element Ai in the array, count the number of element before this element Ai is smaller than it and return count number ar…
Description: Count the number of prime numbers less than a non-negative number, n. 推断一个数是否是质数主要有下面几种方法: 1)直接用该数除于全部小于它的数(非0.1),假设均不能被它整除,则其是质数. 2)除以小于它一半的数.由于大于其一半必然是无法整除.假设均不能被它整除.则其是质数: 3)除以小于sqrt(a)的数,原因例如以下: 除了sqrt(a)之外,其它的两数乘积为a的,一定是比个比sqrt(a)小,…
