20. Valid Parentheses(括号匹配，用桟)

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

The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not.

 

 

 

 

class Solution {
public:
    bool isValid(string s) {
        stack<char> stk;
        for (char c : s) {
            if (c == '(' || c == '[' || c=='{') {
                stk.push(c);
            }  else if (!stk.empty() && leftof(c)==stk.top()) {
                stk.pop();
            } else {
                return false;
            }
        }
        return stk.empty();
    }
    char leftof(char c) {
        if (c == ')') return '(';
        if (c == '}') return '{';
        if (c == ']') return '[';
        return c;
    }
};

 

 

 

 

 

 

 

 

class Solution:
    def isValid(self, s):
        """
        :type s: str
        :rtype: bool
        """
        stack = []
        d={'(':')','[':']','{':'}'}
        for i in s:
            if i in d.keys():#遇到左括号 压桟
                stack.append(i)
            else: #遇到右括号
                if stack ==[]:  #桟为空，没有匹配的
                    return False
                else:
                    if i==d[stack[-1]] : #如果匹配上，弹桟
                        stack.pop()
                    else:
                        return False #没有匹配上
        return stack==[] #如果遍历完之后桟为空，则全部匹配