leetcode20. Valid Parentheses

20. Valid Parentheses

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.

解法一

左半边的括号入栈，右半边的括号需要判断一下是否和栈顶的左半边匹配，如果不匹配，返回false。

tips

String.valueOf()将其他类型转为字符串。
s.toCharArray字符串转char数组。
Stack，泛型里面都是对象类型。字符对象Character和字符判断是否相等 c.equals(‘{‘)。注意是单引号。字符对象有equals方法。
字符比较：c == ‘{‘。

public class Solution {    public boolean isValid(String s) {        Stack<Character> stack = new Stack<Character>();        for (char c : s.toCharArray()) {            if ("{([".contains(String.valueOf(c))) {                stack.push(c);            } else {                if (!stack.empty() && is_valid(c, stack.peek())) {                    stack.pop();                } else {                    return false;                }            }        }        return stack.empty();    }    private boolean is_valid(Character s, Character peek) {        return (s.equals(')') && peek.equals('(')) || (s.equals('}') && peek.equals('{')) || (s.equals(']') && peek.equals('['));    }    private static boolean is_valid2(char s, char peek) {        return s == ')' && peek =='(' || s == '}' && peek == '{' || s == ']' && peek == '[';    }}

解法二

输入’(‘，push’)’。输入’)’,判断栈顶元素是否是’)’。

public class Solution {    public boolean isValid(String s) {        Stack<Character> stack = new Stack<Character>();        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();    }}