Problem 9: Leetcode 96 给你一个整数 n ,求恰由 n 个节点组成且节点值从1 到n 互不相同的 二叉搜索树 有多少种?返回满足题意的二叉搜索树的种数。 比方说如果n = 3,那么输出就是5,五棵树的样子是这样的。 这一个问题看似不像是一个数学题,但是因为它涉及到一个卡特兰数的概念,所以我们拿来说一下...
这是Problem 1的反问题,但这一个题难度要稍微大一些,我们考虑一下怎么做。 这里我们也是考虑使用两个栈来求解。区分于队列,栈是先进后出的,因此每做一次栈的操作都会改变元素的顺序。那么假如说我们有两个栈stack1, stack2。那么一开始我们输入数据的时候,可以把它放到stack1里面,只要我们需要输出了,我们就看看s...
Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity? Credits: Special thanks to@jianchao.li.fighterfor adding this problem and creating all test cases. Subscribeto see which companies asked this question 这题见过,常数空间,线性时间。 第...
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Special thanks to@jianchao.li.fighterfor adding this problem and creating all test cases. x &= -x 是用了树状数组中的lowbit。 1classSolution {2public:3vector<int> singleNumber(vector<int>&nums) {4intx =0;5for(auto n : nums) x ^=n;6inttmp = x, idx =0;7x &= -x;8inta =0,...
1 2 3 4 5 6 class Solution { public: vector<int> maxNumber(vector<int>& nums1, vector<int>& nums2, int k) { } }; 已存储 行1,列 1 运行和提交代码需要登录 Case 1Case 2Case 3 nums1 = [3,4,6,5] nums2 = [9,1,2,5,8,3] k = 5 1 2 3 4 5 6 7 8 9 [3,4,6,...
Dearboy, a goods victualer, now comes to a big problem, and he needs your help. In his sale area there are N shopkeepers (marked from 1 to N) which stocks goods from him.Dearboy has M supply places (marked from 1 to M), each provides K different kinds of goods (marked from 1 ...
题目链接:https://leetcode-cn.com/problems/maximum-number-of-events-that-can-be-attended/ 这道题其实有好多类似的题,都是优先去找最左边的区间,又优先去挑选最右边的端点。所以对于这道题来说,如果现在是第idx天,那么最优的选的肯定是会议结束时间离idx最近的会议,大致思路就是这样,接下来会一点一点的讲解...
Solution to the problem Unit test Note that each of these problemshave passedtheir respective test cases on LeetCode. The unit tests included with each solution in this repo are not comprehensive. They serve as a quick way to drop in a test case, hook up the debugger, and step through th...
You are not suppose to use the library’s sort function for this problem. Solution: 典型的Partition题, 最简单的思路是先对0 和 大于 0 做一次partition, 对1和2做一次partition。这样虽然时间复杂度依然是O(n), 但是还有一...