55：Valid Parentheses

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题目：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.

解题代码一：

// 时间复杂度 O(n)，空间复杂度 O(n)class Solution {public:        bool isValid(string s) {                stack<char> bracket;                for (int i = 0; i != s.size(); ++i) {                        if (s[i] == ')') {                                if (!bracket.empty() && bracket.top() == '(')                                        bracket.pop();                                else return false;                        }                        else if (s[i] == ']') {                                if (!bracket.empty() && bracket.top() == '[')                                        bracket.pop();                                else return false;                        }                        else if (s[i] == '}') {                                if (!bracket.empty() && bracket.top() == '{')                                        bracket.pop();                                else return false;                        }                        else bracket.push(s[i]);                }/*                if (bracket.empty()) return true;                else return false;*/                return bracket.empty();        }};

上面的代码 if… else 语句过多，比较冗杂，如果再考虑其它的括号，比如大括号等等，则会更加冗杂，下面的代码则更为精简

// 时间复杂度 O(n)，空间复杂度 O(n)class Solution {public:        bool isValid(string s) {                string left = "([{";                string right = ")]}";                stack<char> bracket;                for (int i = 0; i != s.size(); ++i) {                        auto index = right.find(s[i]);                        if (index != string::npos) { // s[i] 属于 ')', ']' 或 '}'                                if(!bracket.empty() && bracket.top() == left[index])                                        bracket.pop();                                else return false;                        }                        else bracket.push(s[i]);                }/*                if (bracket.empty()) return true;                else return false;*/                return bracket.empty();        }}

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