https://leetcode.com/problems/count-of-smaller-numbers-after-self/ https://leetcode.com/problems/count-of-smaller-numbers-after-self/discuss/76576/My-simple-AC-Java-Binary-Search-code https://leetcode.com/problems/count-of-smaller-numbers-after-self/discuss/138154/The-C%2B%2B-merge-sort-tem...
1#include <stdio.h>23#include <vector>4classSolution {5public:6std::vector<int> countSmaller(std::vector<int>&nums) {7std::vector<std::pair<int,int> >vec;8std::vector<int>count;9for(inti =0; i < nums.size(); i++){10vec.push_back(std::make_pair(nums[i], i));11count.p...
Count of Smaller Numbers After Self 解法一:Trie树 O((log32)*n ) 解法二: 树状数组 O(nlogn) Leetcode 315. Count of Smaller Numbers After Self 解法一:Trie树 O((log32)*n ) 假设全部元素都是正数,把这些数按二进制位存进Trie数,每个结点都统计以这个点为根树的叶子的数量。 存数的时候从bit...
[leetcode] 315. Count of Smaller Numbers After Self Description 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] O...
public List<Integer> countSmaller(int[] nums) { List<Integer> res = new ArrayList<Integer>(); count = new int[nums.length]; int[] indexes = new int[nums.length]; for(int i = 0; i < nums.length; i++){ indexes[i] = i; ...
LeetCode 315. Count of Smaller Numbers After Self 分治算法练习 本周选择LeetCode第315题Count of Smaller Numbers After Self,难度为hard。 题目描述如下: You are given an integer array nums and you have to return a new counts array. The counts array has the property where coun......
315. Count of Smaller Numbers After Self Total Accepted: 9951 Total Submissions: 32512 Difficulty: Hard 提交网址: 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 ...
我们用 count 来存值出现的次数self.count = 1# 比当前元素值小的元素个数self.smaller = 0# 左子树self.left_child = None# 右子树self.right_child = Noneclass BinarySearchTree:def __init__(self):self.root = Nonedef insert(self, value):if self.root == None:self.root = Node(value)return...
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; while (L <= R) { ...
Leetcode 318 Count of Smaller Numbers After Self (这个题线段树、二分索引树、TreeMap都可以) 动态规划(Dynamic Programming) 基础知识:这里指的是用for循环方式的动态规划,非Memoization Search方式。DP可以在多项式时间复杂度内解决DFS需要指数级别的问题。常见的题目包括找最大最小,找可行性,找总方案数等,一般结...