描述

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.

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

Python

这道题让我们验证输入的字符串是否为括号字符串，包括大括号，中括号和小括号。这里我们需要用一个栈，我们开始遍历输入字符串，如果当前字符为左半边括号时，则将其压入栈中，如果遇到右半边括号时，若此时栈为空，则直接返回false，如不为空，则取出栈顶元素，若为对应的左半边括号，则继续循环，反之返回false

class Solution:
    # @return a boolean
    def isValid(self, s):
        stack = []
        dict = {"]":"[", "}":"{", ")":"("}
        for char in s:
            if char in dict.values():
                stack.append(char)
            elif char in dict.keys():
                if stack == [] or dict[char] != stack.pop():
                    return False
            else:
                return False
        return stack == []

Java

public class Solution {
        public boolean isValid(String s) {
            Stack<Character> stack = new Stack<>();
            Map<Character, Character> map = new HashMap<>();
            map.put('(', ')');
            map.put('[', ']');
            map.put('{', '}');
            for(char c : s.toCharArray()) {
                if(map.containsKey(c)) {
                    stack.push(c);
                } else {
                    if(stack.isEmpty() || map.get(stack.pop()) != c) {
                        return false;
                    }
                }
            }
            return stack.isEmpty();
        }
}