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
这道题看似简单，但是我还是没有一次AC，思路很简答，就是入栈与出栈，遇到左括号入栈，遇见如果有匹配的括号将其出栈，否则返回false，到最后如果栈空则说明括号是合法且匹配的，类似的题目有Generate Parentheses，Longest Valid Parentheses
1 class Solution { 2 public: 3 bool isValid(string s) { 4 stack<char> sta; 5 for (auto c : s) 6 { 7 if (c == '}') 8 { 9 if (sta.empty() || sta.top() != '{') 10 return false; 11 else 12 sta.pop(); 13 } 14 else if (c == ']') 15 { 16 if (sta.empty() || sta.top() != '[') 17 return false; 18 else 19 sta.pop(); 20 } 21 else if (c == ')') 22 { 23 if (sta.empty() || sta.top() != '(') 24 return false; 25 else 26 sta.pop(); 27 } 28 else 29 sta.push(c); 30 } 31 return sta.empty(); 32 } 33 };
有几个地方要注意：1、调用栈的top()函数时一定要先判决栈是否为空，否则会报错；2、多个else if 条件时，一定要用合适的方式进行分类讨论，否则容易重复或者遗漏
以下是一种更简练的写法：
1 class Solution { 2 public: 3 bool isValid(string s) { 4 stack<char> parentheses; 5 for (int i = 0; i < s.size(); ++i) { 6 if (s[i] == '(' || s[i] == '[' || s[i] == '{') parentheses.push(s[i]); 7 else { 8 if (parentheses.empty()) return false; 9 if (s[i] == ')' && parentheses.top() != '(') return false; 10 if (s[i] == ']' && parentheses.top() != '[') return false; 11 if (s[i] == '}' && parentheses.top() != '{') return false; 12 parentheses.pop(); 13 } 14 } 15 return parentheses.empty(); 16 } 17 };