LeetCode 20. Valid Parentheses

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 };

类似的解答请参考：http://www.cnblogs.com/grandyang/p/4424587.html