NO.12 简单 (196. 删除重复的电子邮箱) leetcode-cn.com/problem # 注意是要用到DELETE操作。 DELETE p1 FROM Person p1, Person p2 WHERE p1.Email=p2.Email AND p1.Id>p2.Id NO.13 简单 (1179. 重新格式化部门表) leetcode-cn.com/problem # IF和CASE都可以,同时注意用上SUM函数。 /* select...
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Note: The most water = width * min(left, right), where width = j - i, i and j are the indexes of two sides. We can use two pointers starting from the begging and the end of the array to solve this problem. The core of the problem is how to update the position of two pointer...
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
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...
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
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. ...
triplebee 2018/01/12 6550 leetcode: 68. Text Justification [✗] 其他 Problem # Given an array of words and a length L, format the text such that each line has exactly L characters and is fully (left and right) justified. # # You should pack your words in a greedy approach; that...
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 - Practice Problem 751.IP to CIDR 我请问呢什么题目啊在这里考二进制呢。。。首先我们要会怎样用python 把十进制和二进制互相转换 >>> int("1111", 2) # convert binary to int >>> 15 >>> format(15, '08b') # convert int 15 to binary in 8-bit representation...