NO.15 简单 (197. 上升的温度) leetcode-cn.com/problem # 两表之间相互比较即可 SELECT w1.id FROM Weather w1, Weather w2 WHERE DATEDIFF(w1.recordDate,w2.recordDate)=1 AND w1.Temperature > w2.Temperature NO.16 中等 (626. 换座位) leetcode-cn.com/problem # 关键在于如何将id转换,这里我用...
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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
The encoding rule is:k[encoded_string], where theencoded_stringinside the square brackets is being repeated exactlyktimes. Note thatkis guaranteed to be a positive integer. You may assume that the input string is always valid; No extra white spaces, square brackets are well-formed, etc. Furth...
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.16 Climbing Stairs You are climbing a stair case. It takesnsteps to reach to the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top? classSolution {public:intclimbStairs(intn) {if(n==0)return0;if(n==1)return1;if(n==2)...
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已经存在,则变更其数据值value;如果...
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){ ...
🔥LeetCode solutions in any programming language | 多种编程语言实现 LeetCode、《剑指 Offer(第 2 版)》、《程序员面试金典(第 6 版)》题解 - doocs/leetcode
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 ...