LeetCode in Python 20. Valid Parentheses

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

Solution:

最经典的题之一，遍历字符串放入stack，如果当前字符属于括号的开头，进栈；如果当前字符属于括号的结尾，比较栈顶，不配对返回false，配对出栈，遍历结束栈应该为空，否则返回false。

class Solution(object):
    def isValid(self, s):
        """
        :type s: str
        :rtype: bool
        """
        st = []
        for char in s:
            if char == '(' or char == '[' or char == '{':
                st.append(char)
                continue
            
            if not st:
                return False
            
            if char == ')' and st[-1] != '(':
                return False
            
            if char == ']' and st[-1] != '[':
                return False
            
            if char == '}' and st[-1] != '{':
                return False
            
            st.pop()
            
        return (not st)

还有一种带点trick味道的解法：

class Solution(object):
    def isValid(self, s):
        """
        :type s: str
        :rtype: bool
        """
        while '()' in s or '[]' in s or '{}' in s:
            s = s.replace('()','').replace('[]','').replace('{}','') 
        return (not s)

注意python字符串replace后要赋值，不然原字符串无变化。