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.
这题虽然简单但是我也没有一次就AC，问题在于取top和pop的时候忘了做异常判断，切记切记。
此外找conterpart也许是个比较头疼的问题，但是用hashmap就方便多了
map<char, char> conterpart; bool isValid(string s) { conterpart[‘(‘] = ‘)‘; conterpart[‘[‘] = ‘]‘; conterpart[‘{‘] = ‘}‘; if (s.empty()) return true; stack<char> st; for (int i = 0; i < s.size(); i++) { char c = s.at(i); if (c == ‘(‘ || c == ‘[‘ || c == ‘{‘) { st.push(c); } else if (c == ‘)‘ || c == ‘]‘ || c == ‘}‘){ if (st.empty()) return false; if ( conterpart[st.top()] != c) { return false; } else { st.pop(); } } } if (st.empty()) { return true; } return false; }
原文：http://www.cnblogs.com/agentgamer/p/4099123.html