【LeetCode】Valid Parentheses

版权声明： https://blog.csdn.net/ysq96/article/details/89710060

Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.

An input string is valid if:

1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.

Note that an empty string is also considered valid.

Example 1:

Input: "()"
Output: true

Example 2:

Input: "()[]{}"
Output: true

Example 3:

Input: "(]"
Output: false

Example 4:

Input: "([)]"
Output: false

Example 5:

Input: "{[]}"
Output: true

题意：题目要求给一串可能包含大、中、小括号的字符串，让你判断这个是否正确。括号序列判定就是利用栈啦。

C++：注意当时右括号时要先判断下栈是否为空，若为空则返回false，若不为再进行出栈的操作

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

Python3：最简单的办法其实就是每次将字符串中的正确括号去掉

class Solution:
    def isValid(self, s: str) -> bool:
        while '{}' in s or '[]' in s or '()' in s:
            s=s.replace('{}','')
            s=s.replace('[]','')
            s=s.replace('()','')
        return s==''