Implement a basic calculator to evaluate a simple expression string. The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces . You may assume that the given expression is…

Implement a basic calculator to evaluate a simple expression string. The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero. You may assume that the given ex…

版权声明: 本文为博主Bravo Yeung(知乎UserName同名)的原创文章,欲转载请先私信获博主允许,转载时请附上网址 http://blog.csdn.net/lzuacm. C#版 - Leetcode 10. 正则表达式匹配 - 题解 LeetCode 10. Regular Expression Matching 在线提交 分析 克莱尼星号(算子) 定义及标记法 例子 推广 方法1:递归 方法2:动态规划 在线提交 https://leetcode.com/problems/re…

最近代码写的少了,而leetcode一直想做一个python,c/c++解题报告的专题,c/c++一直是我非常喜欢的,c语言编程练习的重要性体现在linux内核编程以及一些大公司算法上机的要求,python主要为了后序转型数据分析和机器学习,所以今天来做一个难度为hard 的简单正则表达式匹配. 做了很多leetcode题目,我们来总结一下套路: 首先一般是检查输入参数是否正确,然后是处理算法的特殊情况,之后就是实现逻辑,最后就是返回值. 当编程成为一种解决问题的习惯,我们就成为了一名纯粹的程序…

Implement a basic calculator to evaluate a simple expression string. The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces . The expression string contains only non-nega…

Implement a basic calculator to evaluate a simple expression string. The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero. You may assume that the given ex…

Implement a basic calculator to evaluate a simple expression string. The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces . You may assume that the given expression is…

Implement a basic calculator to evaluate a simple expression string. The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces . You may assume that the given expression is…

Implement a basic calculator to evaluate a simple expression string. The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero. You may assume that the given ex…

Implement a basic calculator to evaluate a simple expression string. The expression string contains only non-negativeintegers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero. Example 1: Input: "3+2*2" Ou…

Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'. '.' Matches any single character. '*' Matches zero or more of the preceding element. The matching should cover the entire input string (n…

10. Regular Expression Matching Hard Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'. '.' Matches any single character. '*' Matches zero or more of the preceding element. The matching sh…

10. 正则表达式匹配 给你一个字符串 s 和一个字符规律 p,请你来实现一个支持 '.' 和 '*' 的正则表达式匹配. '.' 匹配任意单个字符 '*' 匹配零个或多个前面的那一个元素 所谓匹配,是要涵盖 整个 字符串 s的,而不是部分字符串. 说明: s 可能为空,且只包含从 a-z 的小写字母. p 可能为空,且只包含从 a-z 的小写字母,以及字符 . 和 *. 示例 1: 输入: s = "aa" p = "a" 输出: false 解释: "…

题目 Implement a basic calculator to evaluate a simple expression string. The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero. You may assume that the given…

https://leetcode.com/problems/regular-expression-matching/ [描述] Implement regular expression matching with support for '.' and '*'. '.' Matches any single character. '*' Matches zero or more of the preceding element. The matching should cover the ent…

这种题都要设置一个符号位的变量 224. Basic Calculator 设置数值和符号两个变量,遇到左括号将数值和符号加进栈中 class Solution { public: int calculate(string s) { stack<int> st; ,flag = ; ;i < s.size();i++){ '){ ; '){ num = num* + flag * (s[i] - '); i++; } res += num; i--; } else if(s[i] ==…

Implement a basic calculator to evaluate a simple expression string. The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces . You may assume that the given expression is…

这道题就可以结合Basic Calculator中的两种做法了,分别是括号运算和四则运算的,则使用stack作为保持的结果,而使用递归来处理括号内的值的. class Solution { public: int calculate(string s) { ,res=,n=s.size(); stack<int> st; char op='+'; ;i<n;i++){ char c=s[i]; '){ num=num*+c-'; }else if(c=='('){ ; for(;i<…

一.题目链接:https://leetcode.com/problems/regular-expression-matching/ 二.题目大意: 实现一个正则表达式,该正则表达式只有两种特殊的字符——“.”和“*”,其中.能表示任意字符,即它可以匹配任意的字符:*表示可以重复前面的字符0次或者多次(例如:a*可以表示成“”(空,相当于重复0次)或者表示成“aaa”重复3次). 三.题解: 这道题目比较常用的方法是递归:该题目的难点之处在于遇到*的时候它可能匹配0次,也可能匹配1次或多次:这种特…

题目链接 https://leetcode.com/problems/regular-expression-matching/?tab=Description   '.' Matches any single character.匹配任何单字符 '*' Matches zero or more of the preceding element.匹配0个或者多个前置元素 采用动态规划方法 public boolean isMatch(String s, String p) 1, If p.char…

这次我觉得我的智商太低,想了很久才写出来.题目是让求镜像二叉树判断,题目如下: Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For example, this binary tree is symmetric: 1 / \ 2 2 / \ / \ 3 4 4 3 But the following is not: 1 / \ 2 2 \ \ 3 3 N…

Basic Calculator II 题目 思路 和这个一样:Basic Calculator I 代码 class ExpressionTransformation { public: string trans_to_postfix_expression_to_s(string); // 将得到的表达式转化为后缀表达式 long long int calculate_from_postfix_expression(); // 依据后缀表达式计算值 private: vector<string…

10. Regular Expression Matching https://www.cnblogs.com/grandyang/p/4461713.html class Solution { public: bool isMatch(string s, string p) { if(p.empty()) return s.empty(); && p[] == '*') )) || (!s.empty() && (s[] == p[] || p[] == ),p)); e…

题目如下: Implement a basic calculator to evaluate a simple expression string. The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces . The expression string contains only no…

这题一年前就做过,当时刚开始刷leetcode,提交了几十次过不去,就放那没管了.今天剑指offer又遇到这题,终于做出来了,用的dp. class Solution { public: bool isMatch(string s, string p) { int s_len=s.size(),p_len=p.size(); vector<vector<,vector<,false)); //dp[i][j]表示s[0,i-1]和p[0,j-1]能否匹配 dp[][]=true;//空串…

每天 3 分钟,走上算法的逆袭之路. 前文合集 每日一道 LeetCode 前文合集 代码仓库 GitHub: https://github.com/meteor1993/LeetCode Gitee: https://gitee.com/inwsy/LeetCode 题目:搜索插入位置 题目来源:https://leetcode-cn.com/problems/search-insert-position/ 给定一个排序数组和一个目标值,在数组中找到目标值,并返回其索引.如果目标值不存在于数组…

Part 1 Introduction The struct of C# program: namespace , class and Main method what is namespace? the namespace declaration, using System, indicates that you are using the System namespace. A namespace is used to organize your code and is collection…

Implement regular expression matching with support for '.' and '*'. DP: public class Solution { public boolean isMatch2(String s, String p) { int starCnt = 0; for (int i = 0; i < p.length(); i++) { if (p.charAt(i) == '*') { starCnt++; } } boolean[] s…

问题描述: Implement regular expression matching with support for '.' and '*'. '.' Matches any single character. '*' Matches zero or more of the preceding element. http://i.cnblogs.com/EditPosts.aspx?opt=1 The matching should cover the entire input string…

题目链接:https://leetcode-cn.com/problems/regular-expression-matching/ 题目描述: 给定一个字符串 (s) 和一个字符模式 (p).实现支持 '.' 和 '*' 的正则表达式匹配. '.' 匹配任意单个字符. '*' 匹配零个或多个前面的元素. 匹配应该覆盖整个字符串 (s) ,而不是部分字符串. 说明: s 可能为空,且只包含从 a-z 的小写字母. p 可能为空,且只包含从 a-z 的小写字母,以及字符 . 和 *. 示例: 示例…