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.
分析：一个很简单的想法是用栈保存左括号。时间复杂度O(n), 空间复杂度O(n)。
class Solution { public: bool isValid(string s) { int n = s.length(); if(n == 0) return true; stack<char> S; for(int i = 0; i < n; i++){ if(s[i] == ‘(‘ || s[i] == ‘{‘ || s[i] == ‘[‘) S.push(s[i]); else if(!S.empty() && match(S.top(), s[i])) S.pop(); else return false; } return S.empty(); } bool match(char l, char r){ return l == ‘(‘ && r == ‘)‘ || l == ‘{‘ && r == ‘}‘ || l == ‘[‘ && r == ‘]‘; } };
原文：http://www.cnblogs.com/Kai-Xing/p/4128698.html