NO.14 简单 (182. 查找重复的电子邮箱) leetcode-cn.com/problem # 相同邮箱但不同Id即为重复数据(记得DISTINCT去重) SELECT DISTINCT p1.Email FROM Person p1, Person p2 WHERE p1.Email=p2.Email AND p1.Id<>p2.Id NO.15 简单 (197. 上升的温度) leetcode-cn.com/problem # 两表之间相互比较...
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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
FindTabBarSize FindBorderBarSize 请你设计并实现一个满足LRU (最近最少使用) 缓存约束的数据结构。 实现LRUCache类: LRUCache(int capacity)以正整数作为容量capacity初始化 LRU 缓存 int get(int key)如果关键字key存在于缓存中,则返回关键字的值,否则返回-1。 void put(int key, int value)如果关键字key已...
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
No longer lazy } if queryLeft = left queryRight = right { // segment completely inside range return st.tree[treeIndex] } if queryLeft midTreeIndex { return st.queryLazyInTree(rightTreeIndex, midTreeIndex+1, right, queryLeft, queryRight) } else if queryRight = midTreeIndex { return ...
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){ ...
These steps should be performed for each problem in our testing set and for each LLM. For the sake of simplicity, there is only one attempt for a LLM to write code on each problem, without any feedback to reiterate to improve the solution. All problems are treated as independent; no con...
今天在使用leetcode做合并区间的问题的时候出现了peek problem no quick fixes available的问题; 问题见下面: for(autoiter= intervals.begin();iter!=intervals.end();iter++) 我使用for语句就会产生问题。 解决方法: vector<vector<int>>::intervals iter;while(iter != intervals.end()) ...
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...