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:
- Open brackets must be closed by the same type of brackets.
- 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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