This is a good question to ask during an interview. For the purpose of this problem, we define empty string as valid palindrome. 【解答】注意大小写,注意数字和字母都要算有效字符。 代码语言:javascript 代码运行次数:0 运行 AI代码解释 public class Solution { public boolean isPalindrome(String s)...
目录 描述 解法一:在一个栈中维持所有元素的出队顺序 思路 入队(push) 出队(pop) 查看队首(peek) 是否为空(empty) Java 实现 Python 实现 解法二:一个栈入,一个栈出 思路 入队(push) 出队(pop) 查看队首(peek) 是否为空(empty) Java 实现 Python 实现 更多 LeetCode 题解笔记可以访问我的 github。
leetcode刷题记录 本文记录一下leetcode刷题记录,记录一下自己的解法和心得。 LeetCode Binary Tree Level Order Traversal II 题目:Binary Tree Level Order Traversal II Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level ...
Related to question Excel Sheet Column TitleGiven a column title as appear in an Excel sheet, return its corresponding column number.For example:A -> 1 B -> 2 C -> 3 ... Z -> 26 AA -> 27 AB -> 28 代码:1 class Solution { 2 public: 3 int titleToNumber(string s) { 4 map...
[LeetCode]43.Multiply Strings 【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. 【分析】 高精度乘法(大数乘法) 其实更多地是考察乘法运算的本质。基本思路是... ...
We can simplify the answer to the Factorial Trailing Zeroes question to the following: (n / 5) + (n / 5^2) + (n / 5^3)... (n / 5^x) We continue until 5^x is greater than n. vartrailingZeroes =function(n) {letnumZeroes =0;for(leti =5; i <= n; i *=5) { ...
给定一个字符串,找到其中第一个不重复的字符,并返回它的下标。如果不存在,则返回-1。在字符串“leetcode”中,第一个不重复的字符是“l”,其下标为0;在字符串“loveleetcode”中,第一个不重复的字符是“v”,其下标为2。 相关知识点: 试题来源: 解析 对于输入字符串的每个字符,统计其出现次数后遍历原字符...
http request: http://ipAddress:Port/SomeResource?Param1=value1&Param2=value2&... so on. This is a http request sample in jmeter that hits a rest api and gets response in JSON format. Here t... Python regular expression question - sub string but not prepended with :) ...
Related to question Excel Sheet Column TitleGiven a column title as appear in an Excel sheet, return its corresponding column number.For example:A -> 1 B -> 2 C -> 3 ... Z -> 26 AA -> 27 AB -> 28代码:1 class Solution { 2 public: 3 int titleToNumber(string s) { 4 int ...