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Given a string containing just the characters ‘(‘, ‘)‘, ‘{‘, ‘}‘, ‘[‘ and ‘]‘, determine if the input string is valid.
The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not.
思路，用栈。之前做过，现在写的更简洁了。 从后向前遍历string,遇到右括号就压栈，遇到匹配的左括号就弹栈，如果遇到不匹配的左括号，那就有错，最后栈空则有效。
class Solution { public: bool isValid(string s) { vector<char> vec; for(int i = s.length() - 1; i >= 0; i--) { if(s[i] == ‘)‘ || s[i] == ‘]‘ || s[i] == ‘}‘) { vec.push_back(s[i]); } else if(s[i] == ‘(‘ && !vec.empty() && vec.back() == ‘)‘) { vec.pop_back(); } else if(s[i] == ‘[‘ && !vec.empty() && vec.back() == ‘]‘) { vec.pop_back(); } else if(s[i] == ‘{‘ && !vec.empty() && vec.back() == ‘}‘) { vec.pop_back(); } else { return false; } } return vec.empty(); } };
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原文地址：http://www.cnblogs.com/dplearning/p/4228659.html
