原题传送门

1. 题目描述

2. Solution 1

1、思路分析
遇到左括号认为优先级低，入栈；遇到右括号，若栈空，不匹配，若栈顶元素不是对应的括号，不匹配，不然弹栈，继续遍历；遍历完，栈不空，不匹配。
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2、代码实现

package Q0099.Q0020ValidParentheses;

import java.util.ArrayDeque;
import java.util.Deque;

public class Solution1 {

    public boolean isValid(String s) {
        Deque<Character> brackets = new ArrayDeque<>();
        for (char c : s.toCharArray()) {
            if (c == '(' || c == '[' || c == '{') brackets.push(c);
            else if (c == ')') {
                if (brackets.isEmpty()) return false;
                else if (brackets.peek() == '(') brackets.pop();
                else return false;
            } else if (c == ']') {
                if (brackets.isEmpty()) return false;
                else if (brackets.peek() == '[') brackets.pop();
                else return false;
            } else if (c == '}') {
                if (brackets.isEmpty()) return false;
                else if (brackets.peek() == '{') brackets.pop();
                else return false;
            }
        }
        return brackets.isEmpty();
    }
}

3. Solution 2

1、思路分析
使用Map保存匹配的括号，简化代码。
2、代码实现

package Q0099.Q0020ValidParentheses;

import java.util.ArrayDeque;
import java.util.Deque;
import java.util.HashMap;
import java.util.Map;

public class Solution2 {

    public boolean isValid(String s) {
        Deque<Character> brackets = new ArrayDeque<>();
        Map<Character, Character> bracketMaps = new HashMap<>();
        bracketMaps.put(')', '(');
        bracketMaps.put(']', '[');
        bracketMaps.put('}', '{');
        for (char c : s.toCharArray()) {
            if (bracketMaps.containsValue(c)) brackets.push(c);  // 左括号 => 进栈
            else if (brackets.isEmpty() || bracketMaps.get(c) != brackets.pop()) return false;
        }
        return brackets.isEmpty();
    }
}

4. Solution 3

1、思路分析
遇到左括号，对应的右括号入栈；遇到右括号，栈空或栈顶元素与当前遍历元素不相等，说明不匹配。
2、代码实现

package Q0099.Q0020ValidParentheses;

import java.util.ArrayDeque;
import java.util.Deque;

public class Solution3 {

    public boolean isValid(String s) {
        Deque<Character> stack = new ArrayDeque<>();
        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();
    }
}

标签：0020,return,brackets,else,Valid,isEmpty,import,false,LeetCode
来源： https://www.cnblogs.com/junstat/p/15983399.html