3)遍历结束的时候,假如栈不空,表示不配对,否者配对成功。 建议和leetcode 32. Longest Valid Parentheses 最长有效括号长度和 leetcode 678. Valid Parenthesis String 有效括号的判断 一起学习 代码如下: import java.util.ArrayList; import java.util.List; public class Solution { public boolean isValid(String...
Any right parenthesis')'must have a corresponding left parenthesis'('. Left parenthesis'('must go before the corresponding right parenthesis')'. '*'could be treated as a single right parenthesis')'or a single left parenthesis'('or an empty string. An empty string is also valid. Example 1:...
但是Leetcode的solution十分优秀,它的做法是这样的,定义一个贡献,是目前多余的左括号数量,那么可以定义星号作为左括号和右括号的情况下,会有一个上限和下限,上限表示星号全部是左括号的情况,下限表示星号全部是右括号的情况;当碰到左括号时,上限和下限都得加1;碰到右括号时,上限减一,下限减一,但是,如果上限大于等于...
【leetcode】Valid Parenthesis String 题目: Given a string containing only three types of characters: '(', ')' and '*', write a function to check whether this string is valid. We define the validity of a string by these rules: 1.Any left parenthesis '(' must have a corresponding right...
Having only one of the parenthesis: ) public class Solution { public bool IsValid(string s) { Stack < char > paranthesisStack = new(); foreach(char item in s) { switch (item) { case '{': paranthesisStack.Push('}'); break; case '[': paranthesisStack.Push(']'); break; case ...
In one move, you can insert a parenthesis at any position of the string. For example, if s = "()))", you can insert an opening parenthesis to be "(()))" or a closing parenthesis to be "()))". Return the minimum number of moves required to make s valid. Example 1: Input: ...
Poblem Statement: Given a string containing just the characters '(', ')', '{', '}', '[', and ']', determine if the input string is valid. An input string is valid if: Open brackets must be closed by the same type of brackets. Open bracke...
publicclassSolution{publicintlongestValidParentheses(String s){if(s==null||s.length()==0)return0;Stack<Integer>stack=newStack<>();intmaxLen=0;intaccumulatedLen=0;for(inti=0;i<s.length();i++){if(s.charAt(i)=='(')stack.push(i);else{if(stack.isEmpty())accumulatedLen=0;//())(...
File metadata and controls Code Blame 58 lines (49 loc) · 943 Bytes Raw #include<iostream> #include<stack> using namespace std; bool valid_parenthesis(string s){ int n=s.length(); stack<char> st; bool ans=true; for(int i=0;i<n;i++){ if(s[i]=='{' or s[i]=='(' or...
https://leetcode.com/problems/valid-parenthesis-string/discuss/139759/Java-Very-easy-solution.-No-recursion-dp. https://leetcode.com/problems/valid-parenthesis-string/discuss/107577/Short-Java-O(n)-time-O(1)-space-one-pass https://leetcode.com/problems/valid-parenthesis-string/discuss/107572/...