LeetCode: 20. Valid Parentheses

0509第1题（虽然是08做的，但这会已经09了）

题目

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 brackets must be closed in the correct order.
Note that an empty string is also considered valid.

思路

直接用一个后进先出的stack来匹配就好

class Solution {
public:bool isValid(string s) {stack<char> mystack;if(s.length()==1)return false;for(int i=0; i<s.length(); i++){if(mystack.size()==0 & (s[i]==')' || s[i]=='}' || s[i]==']'))return false;else if(s[i]=='(' || s[i]=='{' || s[i]=='[')mystack.push(s[i]);else if(s[i]=='}' & mystack.top()=='{' & mystack.size()!=0)mystack.pop();else if(s[i]==']' & mystack.top()=='[' & mystack.size()!=0)mystack.pop();else if(s[i]==')' & mystack.top()=='(' & mystack.size()!=0)mystack.pop();else return false;}if(mystack.size()==0)return true;else return false;      }
};

然后在提交过程中，发现自己老是忘了各种特殊样例，还是要多刷题鸭！
加油，你是最胖的！

刚刚去偷瞄了大佬的答案，虽然runtime和memory和我差不多，但素人家的看起来很简洁啊！

class Solution {
public:bool isValid(string s) {stack<char> paren;for (char& c : s) {switch (c) {case '(': case '{': case '[': paren.push(c); break;case ')': if (paren.empty() || paren.top()!='(') return false; else paren.pop(); break;case '}': if (paren.empty() || paren.top()!='{') return false; else paren.pop(); break;case ']': if (paren.empty() || paren.top()!='[') return false; else paren.pop(); break;default: ; // pass}}return paren.empty() ;}
};

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LeetCode: 20. Valid Parentheses

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