LeetCode-Valid Parentheses

Description：
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. Open
2. 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：根据所给的字符串，包含’(‘，’)’，’{‘， ‘}’， ‘[’ 和 ‘]’这三种括号，要求判断其是否合理；即左括号和右括号匹配，并且顺序也要求正确；

解法：可以使用一个栈来实现这个判断，我们将左括号入栈，一旦遇到右括号，弹出栈顶的符号，判断是否是与其相匹配的另外一半；

class Solution {
    public boolean isValid(String s) {
        if(s.length() == 0){
            return true;
        }//空串
        Stack<Character> brackets = new Stack<>();
        for(int i=0; i<s.length(); i++){
            char ch = s.charAt(i);
            switch(ch){
                case ')':
                    if(brackets.isEmpty() || brackets.pop() != '(') return false;
                    break;
                case '}':
                    if(brackets.isEmpty() || brackets.pop() != '{') return false;
                    break;
                case ']':
                    if(brackets.isEmpty() || brackets.pop() != '[') return false;
                    break;
                default:
                    brackets.push(ch);//左半部分入栈
                    break;
            }
        }
        return brackets.isEmpty();//栈为空，说明正确匹配
    }
}