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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：利用栈来判断配对的括号，如果都可以配对成功，最后栈的元素数量应该为0。题目给定的字符串只包含括号，所以每次遇到是(‘, ‘{‘或者‘[‘，直接入栈；遇到‘]‘,‘)‘或者‘]‘，只有当栈不为空&&栈顶元素与之配对时，才弹出栈，否则直接返回false。
代码：
1 class Solution { 2 public: 3 bool isValid(string s) { 4 stack<char> cs; 5 unordered_map<char, char> um; 6 um[‘(‘] = ‘)‘; 7 um[‘{‘] = ‘}‘; 8 um[‘[‘] = ‘]‘; 9 for (int i = 0; i < s.size(); i++) { 10 if (s[i] == ‘(‘ || s[i] == ‘{‘ || s[i] == ‘[‘) { 11 cs.push(s[i]); 12 continue; 13 } 14 if (!cs.empty() && um[cs.top()] == s[i]){ 15 cs.pop(); 16 continue; 17 } 18 return false; 19 } 20 return cs.empty(); 21 } 22 };
Leetcode 20. Valid Parentheses
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原文地址：http://www.cnblogs.com/Deribs4/p/5838714.html
