leetcode: 20. Valid Parentheses

题目：

Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.

An input string is valid if:

Open brackets must be closed by the same type of brackets.
Open brackets must be closed in the correct order.
Note that an empty string is also considered valid.

Example 1:

Input: "()"
Output: true
Example 2:

Input: "()[]{}"
Output: true
Example 3:

Input: "(]"
Output: false
Example 4:

Input: "([)]"
Output: false
Example 5:

Input: "{[]}"
Output: true

分析：这个题目比较简单，只是怎样能够把代码写得更简洁，我的结果为

还不错吧，空间较大是因为使用了 map，为了使代码更简单。
循环使用 for(char &c : s) 会使代码更简洁。
虽然写出来了，但写得还是不够好，修改了几次，刚开始的代码更加冗余，这是思路不清晰的结果，还需勤加练习。

 bool isValid(string s) {
        stack<int> st;
        map<char, char> maps;
        maps[')'] = '(';  maps[']'] = '['; maps['}'] = '{';
            
        int i = 0;
        while(i < s.length()){         
            if(s[i] == '}' || s[i] == ')' || s[i] == ']'){
                if(!st.empty() && maps[s[i]] == st.top()){
                    st.pop();
                    ++i;
                }
                else{
                    return false;
                }
            }
            else{
                st.push(s[i]);
                ++i;
            }    
        }
        if(!st.empty()){
            return false;
        }
        return true;        
    }