leecode-20. 有效的括号

  • 给定一个只包括 ‘(‘，’)’，’{‘，’}’，’[‘，’]’ 的字符串，判断字符串是否有效。

  • 有效字符串需满足：

    • 左括号必须用相同类型的右括号闭合。
    • 左括号必须以正确的顺序闭合。
    • 注意空字符串可被认为是有效字符串。
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示例 1:
输入: "()"
输出: true
示例 2:

输入: "()[]{}"
输出: true
示例 3:

输入: "(]"
输出: false
示例 4:

输入: "([)]"
输出: false
示例 5:

输入: "{[]}"
输出: true

解法一：

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class Solution:
    def isValid(self, s):
        stack = []
        for st in s:
            if stack and stack[-1]+st in ["()","{}","[]"]:
                stack.pop()
                continue
            stack.append(st)
        return len(stack) == 0

解法二

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class Solution:
    def isValid(self, s):
        dic = {'{': '}', '[': ']', '(': ')', '?': '?'}
        stack = ['?']
        for c in s:
            if c in dic:
                stack.append(c)
            elif dic[stack.pop()] != c:
                return False
        return len(stack) == 1

# 作者：jyd
# 链接：https: // leetcode - cn.com / problems / valid - parentheses / solution / valid - parentheses - fu - zhu - zhan - fa - by - jin407891080 /


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class Solution {
    public boolean isValid(String s) {
        int n = s.length();
        if (n % 2 == 1) {
            return false;
        }

        Map<Character, Character> pairs = new HashMap<Character, Character>() {{
            put(')', '(');
            put(']', '[');
            put('}', '{');
        }};
        Deque<Character> stack = new LinkedList<Character>();
        for (int i = 0; i < n; i++) {
            char ch = s.charAt(i);
            if (pairs.containsKey(ch)) {
                if (stack.isEmpty() || stack.peek() != pairs.get(ch)) {
                    return false;
                }
                stack.pop();
            } else {
                stack.push(ch);
            }
        }
        return stack.isEmpty();
    }
}

//作者：LeetCode-Solution
//链接：https://leetcode-cn.com/problems/valid-parentheses/solution/you-xiao-de-gua-hao-by-leetcode-solution/