Count of Smaller Number before itself

【Count of Smaller Number before itself】的更多相关文章

LeetCode "Count of Smaller Number After Self"

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;…

Lintcode249 Count of Smaller Number before itself solution 题解

[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…

Lintcode: Count of Smaller Number

Give you an integer array (index from 0 to n-1, where n is the size of this array, value from 0 to 10000) and an query list. For each query, give you an integer, return the number of element in the array that are smaller than the given integer. Have…

LintCode "Count of Smaller Number before itself"

Warning: input could be > 10000... Solution by segment tree: struct Node { Node(), left(nullptr), right(nullptr) {}; int start; int end; int cnt; // Node *left; Node *right; }; class Solution { Node *pRoot; void update(int v) { Node *p = pRoot; while…

Count of Smaller Number before itself

Give you an integer array (index from 0 to n-1, where n is the size of this array, value from 0 to 10000) . For each element Ai in the array, count the number of element before this elementAi is smaller than it and return count number array. Example…

Lintcode248 Count of Smaller Number solution 题解

[题目描述] Give you an integer array (index from 0 to n-1, where n is the size of this array, value from 0 to 10000) and an query list. For each query, give you an integer, return the number of element in the array that are smaller than the given integer…

[Swift]LeetCode315. 计算右侧小于当前元素的个数 | Count of Smaller Numbers After Self

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…

【Count of Smaller Number before itself】的更多相关文章

LeetCode "Count of Smaller Number After Self"

Lintcode249 Count of Smaller Number before itself solution 题解

Lintcode: Count of Smaller Number

LintCode "Count of Smaller Number before itself"

Count of Smaller Number before itself

Lintcode248 Count of Smaller Number solution 题解

[Swift]LeetCode315. 计算右侧小于当前元素的个数 | Count of Smaller Numbers After Self

leetcode 315. Count of Smaller Numbers After Self 两种思路（欢迎探讨更优解法）

[LeetCode] 315. Count of Smaller Numbers After Self (Hard)

leetcode 315. Count of Smaller Numbers After Self 两种思路