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The problem: Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is defined as follows: The left subtree of a node contains only nodes with keys less than the node's key. The right subtree of a node contains only nodes with keys greater than the ...
53. 最大子数组和 - 给你一个整数数组 nums ,请你找出一个具有最大和的连续子数组(子数组最少包含一个元素),返回其最大和。 子数组 是数组中的一个连续部分。 示例 1: 输入:nums = [-2,1,-3,4,-1,2,1,-5,4] 输出:6 解释:连续子数组 [4,-1,2,1] 的和最大,为
Can you solve this real interview question? Text Justification - Given an array of strings words and a width maxWidth, format the text such that each line has exactly maxWidth characters and is fully (left and right) justified. You should pack your word
给你一个二叉树的根节点root, 检查它是否轴对称。 示例1: 输入:root = [1,2,2,3,4,4,3]输出:true 示例2: 输入:root = [1,2,2,null,3,null,3]输出:false 提示: 树中节点数目在范围[1, 1000]内 -100 <= Node.val <= 100 进阶:你可以运用递归和迭代两种方法解决这个问题吗?
// Solution 1: There's currently no known formula for N-queen problem. So we'll do it the old way. 代码1 //Code 1 53 Maximum Subarray // #53 最大子数组和 描述:如题。 //#53Description: Maximum Subarray | LeetCode OJ 解法1:教材经典例子。
The Skyline Problem 219. Contains Duplicate II 220. Contains Duplicate III 222. Count Complete Tree Nodes 223. Rectangle Area 224. Basic Calculator 225. Implement Stack using Queues 226. Invert Binary Tree 229. Majority Element II 230. Kth Smallest Element in a BST 231. Power of Two 232. ...
testHead=testHead.next; } }if(!valid){break; }if(pre ==null){ pre=testHead; ret=pre; }else{ pre.next=testHead; } ListNode oldPre=current; ListNode next=current.next; current.next=testHead.next; pre=current; current=next;for(inti = 1; i < k; ++i){ ...
第二是problem solving,要展示自己的解决问题的策略。比如选数据结构什么的,一定要讲明白为什么要用各种算法和数据结构。 第三是coding。在前面两步的基础上,写出来干净正确的代码。 第四是testing,一定要回去验证自己的代码。这个过程中可以分析一下代码的复杂度。 怎么去练以上的步骤,有详细过程解析的入门教程就能...
Leetcode#ProblemLevelTagsTimeSpaceLanguageSequence N/A Jump Game II.java Hard [Array, Coordinate DP, DP, Greedy] O(n) O(1) Java 0 N/A Majority Number II.java Medium [Enumeration, Greedy] Java 1 N/A Search a 2D Matrix II.java Medium [Binary Search, Divide and Conquer] Java 2 N/A...