LeetCode | #20 Valid Parentheses

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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.

思路：

“([])”这样的形式也是valid，用栈，后进先出，判断规则如下：

如果是'(', '[', '{'，就直接入栈；
如果是')', ']', '}'，就判断栈顶和该字符是不是一对，是就将栈顶pop，继续1，否则就返回false；
如果最后栈不为空，返回false，否则返回true；

可以用map也可以不用

import java.util.HashMap;import java.util.Map;import java.util.Stack;public class ValidParentheses {/*//用map存对应关系public boolean isValid(String s) {Map<Character, Character> map = new HashMap<Character, Character>(){{put(')', '(');put(']', '[');put('}', '{');}};Stack<String> stack = new Stack<>();int n = s.length();for(int i=0; i<n; i++){if(stack.empty()){stack.push(s.charAt(i));continue;}if(stack.peek() == (map.get(s.charAt(i)))){stack.pop();}else{stack.push(s.charAt(i));}}if(stack.empty())return true;return false;    }*/public boolean isValid(String s) {Stack<Character> stack = new Stack<>();int n = s.length();char c;for(int i=0; i<n; i++){c = s.charAt(i);if(c==')' || c==']' || c=='}'){if(stack.empty()) return false;char c2 = stack.peek();if((c==')' && c2!='(') || (c==']' && c2!='[') || (c=='}' && c2!='{')){return false;}stack.pop();}else{stack.push(c);}}if(stack.empty())return true;return false;    }}

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