原題傳送門

1. 題目描述

2. Solution 1

1、思路分析
遇到左括號認為優先順序低，入棧；遇到右括號，若棧空，不匹配，若棧頂元素不是對應的括號，不匹配，不然彈棧，繼續遍歷；遍歷完，棧不空，不匹配。
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2、程式碼實現

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儲存匹配的括號，簡化程式碼。
2、程式碼實現

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);  // 左括號 => 進棧
            else if (brackets.isEmpty() || bracketMaps.get(c) != brackets.pop()) return false;
        }
        return brackets.isEmpty();
    }
}
 

4. Solution 3

1、思路分析
遇到左括號，對應的右括號入棧；遇到右括號，棧空或棧頂元素與當前遍歷元素不相等，說明不匹配。
2、程式碼實現

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();
    }
}