[LeetCode] Valid parentheses

最新推荐文章于 2025-12-04 16:02:12 发布

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本文介绍了一种使用栈数据结构来验证括号是否正确配对的方法。通过Java和Ruby两种语言实现，确保输入字符串中的括号（包括圆括号、方括号及花括号）按正确的顺序闭合。

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.

第一想法果断是用栈啊~~

java code

public class Solution {
    public boolean isValid(String s) {
        int length = s.length();
        if(length == 0 || length%2 == 1) return false;
        String record = "(){}[]";
        Stack<Character> myStack = new Stack<Character>();
        if(s.charAt(0) == '(' || s.charAt(0) == '{' || s.charAt(0) == '[') myStack.push(s.charAt(0));
        else return false;
        for(int i=1; i<length; i++) {
            if(s.charAt(i) == '(' || s.charAt(i) == '{' || s.charAt(i) == '[') myStack.push(s.charAt(i));
            else {
                int pos = record.indexOf(s.charAt(i));
                if(record.charAt(pos-1) != myStack.peek()) return false;
                else {
                    myStack.pop();
                }
            }
        }
        if (myStack.empty()) return true;
        else return false;
    }
}

243 ms

ruby code

# @param {String} s
# @return {Boolean}
def is_valid(s)
    length = s.length
    return false if (length == 0 || length%2 == 1)
    left = ['(', '{', '[']
    right = [')', '}', ']']
    record = []
    length.times do |i|
        if left.include?(s[i])
            record.push(s[i])
        else
            if(left[right.index(s[i])] != record[-1])
                return false
            else
                record.pop
            end
        end
    end
    if record.empty?
        return true
    else 
        return false
    end
end