C++实现LeetCode(20.验证括号)

[LeetCode] 20. Valid Parentheses 验证括号

Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.

An input string is valid if:

  1. Open brackets must be closed by the same type of brackets.
  2. Open brackets must be closed in the correct order.

Note that an empty string is also considered valid.

Example 1:

Input: "()"
Output: true

Example 2:

Input: "()[]{}"
Output: true

Example 3:

Input: "(]"
Output: false

Example 4:

Input: "([)]"
Output: false

Example 5:

Input: "{[]}"
Output: true

这道题让我们验证输入的字符串是否为括号字符串，包括大括号，中括号和小括号。这里需要用一个栈，开始遍历输入字符串，如果当前字符为左半边括号时，则将其压入栈中，如果遇到右半边括号时，若此时栈为空，则直接返回 false，如不为空，则取出栈顶元素，若为对应的左半边括号，则继续循环，反之返回 false，代码如下：

 方法一：

class Solution {
public:
    bool isValid(string s) {
        stack<char> parentheses;
        for (int i = 0; i < s.size(); ++i) {
            if (s[i] == '(' || s[i] == '[' || s[i] == '{') parentheses.push(s[i]);
            else {
                if (parentheses.empty()) return false;
                if (s[i] == ')' && parentheses.top() != '(') return false;
                if (s[i] == ']' && parentheses.top() != '[') return false;
                if (s[i] == '}' && parentheses.top() != '{') return false;
                parentheses.pop();
            }
        }
        return parentheses.empty();
    }
};

方法二：

class Solution {
public:
    bool isValid(string s) {
        int n = s.size();
        if (n % 2 == 1) {
            return false;
        }

        unordered_map<char, char> pairs = {
            {')', '('},
            {']', '['},
            {'}', '{'}
        };
        stack<char> stk;
        for (char ch: s) {
            if (pairs.count(ch)) {
                if (stk.empty() || stk.top() != pairs[ch]) {
                    return false;
                }
                stk.pop();
            }
            else {
                stk.push(ch);
            }
        }
        return stk.empty();
    }
};

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