【LeetCode20】【Valid Parentheses】

题目：

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.

就是说要判断内容为”{“,”}”,”[“,”]”,”(“,”)”的字符串，是否一一对应，有开始有结束。

代码

1.个人思路

/**我优化几遍后的代码
     * 最短用时:8ms
     * beats:93.67%
     */
    public boolean isValid1(String s) {
        if(s.length()%2!=0||s.length()==0){return false;}   
        Stack stack = new Stack();
        for(char c:s.toCharArray()){
            if(c=='('||c=='{'||c=='['){stack.push(c);}                  
            else if(stack.isEmpty()){return false;}         
            else if(c==']'){if(stack.peek()!='['){return false;}stack.pop();}                               
            else if(c==')'){if(stack.peek()!='('){return false;}stack.pop();}                               
            else if(c=='}'){if(stack.peek()!='{'){return false;}stack.pop();}                               
        }                   
        return stack.isEmpty();
    }

2.网上大神思路

/**
     * 网上大神思路
     * 用时：10ms
     * beats:65.31%
     */ 
    public boolean isValid2(String s){
        Stack stack = new Stack();
        for(char c:s.toCharArray()){
            if(c=='('){stack.push(')');}
            else if(c=='{'){stack.push('}');}
            else if(c=='['){stack.push(']');}
            else if(stack.isEmpty()||stack.pop()!=c){return false;}
        }
        return stack.isEmpty();
    }