【LeetCode-面试算法经典-Java实现】【020-Valid Parentheses(括号验证)】

【020-Valid Parentheses(括号验证)】


【LeetCode-面试算法经典-Java实现】【所有题目目录索引】

原题

  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.

题目大意

  给定一个只包含(‘, ‘)’, ‘{‘, ‘}’, ‘[’ 和‘]’的字符串,验证它是否是有效的。括号必须配对,并且要以正确的顺序。

解题思路

  用一个栈来对输入的括号串进行处理,如果是左括号就入栈,如果是右括号就与栈顶元素看是否组成一对括号,组成就弹出,并且处理下一个输入的括号,如果不匹配就直接返回结果。

代码实现

import java.util.*;

public class Solution {
    public boolean isValid(String s) {

        Deque<Character> stack = new LinkedList<>();
        int index = 0;
        Character top;
        while (index < s.length()) {
            Character c = s.charAt(index);
            switch (c) {
                case ‘(‘:
                case ‘[‘:
                case ‘{‘:
                    stack.addFirst(c);
                    break;
                case ‘)‘:

                    if (stack.isEmpty()) {
                        return false;
                    }

                    top = stack.getFirst();
                    if (top == ‘(‘) {
                        stack.removeFirst();
                    } else if (top == ‘[‘ || top == ‘{‘) {
                        return false;
                    } else {
                        stack.addFirst(c);
                    }
                    break;
                case ‘]‘:

                    if (stack.isEmpty()) {
                        return false;
                    }

                    top = stack.getFirst();
                    if (top == ‘[‘) {
                        stack.removeFirst();
                    } else if (top == ‘(‘ || top == ‘{‘) {
                        return false;
                    } else {
                        stack.addFirst(c);
                    }
                    break;
                case ‘}‘:

                    if (stack.isEmpty()) {
                        return false;
                    }

                    top = stack.getFirst();
                    if (top == ‘{‘) {
                        stack.removeFirst();
                    } else if (top == ‘[‘ || top == ‘(‘) {
                        return false;
                    } else {
                        stack.addFirst(c);
                    }
                    break;
                default:
                    return false;
            }

            index++;
        }

        return stack.isEmpty();
    }
}

评测结果

  点击图片,鼠标不释放,拖动一段位置,释放后在新的窗口中查看完整图片。

特别说明

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时间: 2024-10-25 15:31:22

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