《leetcode》valid-parentheses

题目描述

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.

解析：利用栈进行匹配，详情见代码注解

     public static boolean isValid(String s) {         Stack<Character> stack = new Stack<>();         Character temp;         for(int i=0;i<s.length();i++){             temp=s.charAt(i);             if(temp=='('||temp=='['||temp=='{'){//当前字符是左括号，入栈                 stack.push(temp);             }else {                 if(stack.isEmpty()){//当前括号是右括号，栈空说明没有左括号                     return false;                 }                 Character top=stack.peek();//取栈顶元素                 if(temp==')'){//需要判断当前元素与栈顶元素的匹配情况，例如："([)]"   返回：false                     if(top=='('){                         stack.pop();                     }                 }                 else if(temp==']'){                     if(top=='['){                         stack.pop();                     }                 }else {                     stack.pop();                 }             }         }         return stack.isEmpty();//栈空则说明全部匹配了     }