有效括号，力扣20

最新推荐文章于 2025-12-17 15:36:40 发布

原创最新推荐文章于 2025-12-17 15:36:40 发布 · 242 阅读

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文章标签：

#python #leetcode #c#

本文对比分析了C#和Python实现的检查字符串中有效括号的算法，展示了两种语言中如何利用列表和索引来确保括号配对的正确性。

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题目链接。有效括号

C#
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   public static bool IsValid(string s)
        {
            List<char> left_list = new List<char>() { '(', '{', '[' };
            List<char> right_list = new List<char>() { ')', '}', ']' };
            if (!left_list.Contains(s[0]))
            {
                return false;
            }
            if (s.Length%2 !=0)
            {
                return false;
            }
            int target = 0;
            int index = 0;
            List<char> next_list = new List<char>();
            foreach (char item in s)
            {
                index = 0;
                if (target<=1 || left_list.Contains(item))
                {
                    if (!left_list.Contains(item))
                    {
                        return false;
                    }
                    index = left_list.IndexOf(item);
                    next_list.Add(right_list[index]);
                    target = 2;
                    
                }
                else
                {
                    if (item ==next_list.Last())
                    {
                        next_list.RemoveAt(next_list.Count - 1);
                    }
                    else
                    {
                        return false;
                    }
                    if (next_list.Count <= 0)
                    {
                        target = 1;
                    }
                }
            }
            if (next_list.Count>0)
            {
                return false;
            }
            return true;
        }

python2:
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def isValid(s):
    left_list = ['(', '{', '[']
    right_list = [')', '}', ']']
    if not left_list.__contains__(s[0]):
        return False
    if not (len(s) % 2) == 0:
        return False
    target = 0
    next_list = []
    for ss in s:
        if target <= 1:
            if not left_list.__contains__(ss):
                return False
            index = left_list.index(ss)
            next_list.append(right_list[index])
            target = 2
        else:
            if left_list.__contains__(ss):
                index = left_list.index(ss)
                next_list.append(right_list[index])
                target = 2
            elif ss == next_list[-1]:
                next_list.pop()
            else:
                return False
            if len(next_list) <= 0:
                target = 1
    if len(next_list) > 0:
        return False
    return True