自动合并测试LeetCode解题方法 在leetcode.com上答题,Run Code或者Sumbmit通常要Spending一会,如果提交一次就Accepted那还好,如果反复Wrong Answer,很耽误时间.为了调高效率和减少挫折(来回提交,一直Wrong Answer倍受打击),我采取在本地Jupyter notebook上coding,测试通过后再提交的方式.本篇主要介绍自动测试的方法. 1. LeetCode模式 打开leetcode一道题后,左侧有一个Description选项

题目描述: 给定一个二叉树,检查它是否是镜像对称的. 例如,二叉树 [1,2,2,3,4,4,3] 是对称的. 1 / \ 2 2 / \ / \ 3 4 4 3 但是下面这个 [1,2,2,null,3,null,3] 则不是镜像对称的: 1 / \ 2 2 \ \ 3 3 说明: 如果你可以运用递归和迭代两种方法解决这个问题,会很加分. 思路解析: 思路一:对二叉树进行中序遍历,判断节点值是否对称分布(LeetCode测试用例未通过,但本地测试通过) public static boolea

1.把一个字符串转换成整数——<剑指Offer>P29 2.求链表中的倒数第k个结点——<剑指Offer>P30 3.实现Singleton模式——<剑指Offer>P48 参考:http://www.cnblogs.com/tonglin0325/p/5196818.html 4.数组的内存是连续的,所以数组的时间效率很高,可以用来实现简单的哈希表——<剑指Offer>35题“第一个只出现一次的字母” 5.面试题3:二维数组中的查找——<剑指Offe

LeetCode 跟树结构相关的题目的测试用例中大多是通过String数组来构造树.例如{2,#,3,#,4,#,5,#,6},可以构造出如下的树(将树结构逆时针选择90度显示): 6            5        4    32 很直观地可以理解,输入的String数组是对树结构进行“层序”遍历得到的结果.以下代码用于构造树结构,并提供printTree用于打印树结构. package util; import java.util.LinkedList; import java.ut

121. Best Time to Buy and Sell Stock Say you have an array for which the ith element is the price of a given stock on day i. If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algor

设计并实现最近最久未使用(Least Recently Used)缓存. 题目描述: Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and set. get(key) - Get the value (will always be positive) of the key if the key ex

LeetCode相关的网上资源比较多,看到题目一定要自己做一遍,然后去学习参考其他的解法. 链接: https://oj.leetcode.com/problems/min-stack/ 题目描述: Design a stack that supports push, pop, top, and retrieving the minimum element in constant time. push(x) -- Push element x onto stack. pop() -- Remov

Given a linked list, return the node where the cycle begins. If there is no cycle, returnnull. Follow up: Can you solve it without using extra space? 解题思想: 用两个指针来操作本题,一个慢指针slow,一个快指针fast,慢指针一下走一步,慢指针一下走两不,如果不含环,那么当快指针到尾部的时候 return null,如果有环,那么一定存在一点

(1)Best Time to Buy and Sell Stock Total Accepted: 10430 Total Submissions: 33800My Submissions Say you have an array for which the ith element is the price of a given stock on day i. If you were only permitted to complete at most one transaction (ie

(1)Linked List Cycle Total Accepted: 13297 Total Submissions: 38411 Given a linked list, determine if it has a cycle in it.   Follow up: Can you solve it without using extra space? (url:http://oj.leetcode.com/problems/linked-list-cycle/) 这一题是给定一个link

Missing Number Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array. For example, Given nums = [0, 1, 3] return 2. Note: Your algorithm should run in linear runtime complexity. Could you

题目地址:https://oj.leetcode.com/problems/two-sum/ Two Sum Given an array of integers, find two numbers such that they add up to a specific target number. The function twoSum should return indices of the two numbers such that they add up to the target, w

1 题目 Contains Duplicate I Given an array of integers, find if the array contains any duplicates. Your function should return true if any value appears at least twice in the array, and it should return false if every element is distinct. 接口: public bo

1 题目 Given two numbers represented as strings, return multiplication of the numbers as a string. Note: The numbers can be arbitrarily large and are non-negative. 接口: public String multiply(String num1, String num2); 2 思路 大整数的乘法,不能够简单用Integer.valueOf(

题目链接:https://leetcode.com/problems/happy-number/ 题目理解:实现isHappy函数,判断一个正整数是否为happy数 happy数:计算要判断的数的每一位的平方和,平方和为1则为happy数,不为1则将原数替换为此平方和,继续上述步骤,直到等于1(为happy数),或者进入不包含1的无限循环(非happy数). 测试用例:19为happy数: 解题思路: 定义set集合存每次计算得到的需判断数n,循环计算新的原数n,终止条件为n等于1或者set中出

You are climbing a stair case. It takes n steps 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? 一共有n阶楼梯,每次只能爬1阶或2阶,问爬到顶端一共有多少种方法 方法一: 利用二叉树的思想,对于n=3时有3种方法,有几个叶子节点便有几种方法 void climb(

问题 给出一个三角形,找出从顶部至底部的最小路径和.每一步你只能移动到下一行的邻接数字. 例如,给出如下三角形: [ [2], [3,4], [6,5,7], [4,1,8,3] ] 从顶部至底部的最小路径和为11(即2+3+5+1=11). 注意: 加分项-如果你能只使用O(n)的额外空间,n为三角形中的总行数. 初始思路 最直接的思路就是把路径都走一遍.即从顶点出发,分别往左中右移动(如果可能的话):然后对走到的位置继续进行同样移动,直到走到最后一行.这样就可以得到一个递归的方案,而递归的结

本人是研二程旭猿一枚,还有半年多就要找工作了,想想上一年度面试阿里的算法工程师挂了,心有不甘啊,主要还是准备不足,对一些常见的算法问题没有去组织准备,为了明年找一份好的实习,就从现在开始,好好准备吧,每天刷一道题. LeetCode,OJ系统,不用多说,国外的,题目也非常好,只有精选的152道,不像别的OJ系统有刷不完的题,而且题目有浅有深,测试用例也非常严格,并且对时间复杂.空间复杂度都有一定的要求,非常考验一个程序猿的思维完整性. 本栏目记录我对LeetCode题目的解题思路与完整程序,程序

Given an array of integers, find two numbers such that they add up to a specific target number. The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that

题目Median of Two Sorted Arrays(难度Hard) 方案1,数组合并&排序调用Java方法 import java.util.* class Solution { fun findMedianSortedArrays(nums1: IntArray, nums2: IntArray): Double { val lenNums1 = nums1.size val lenNums2 = nums2.size val array = Arrays.copyOf(nums1,

题目难度:Medium 题目: Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target. Note: The solution set must not contain duplicate quadrup