246. Strobogrammatic Number

I did naive way first.

1:  class Solution {  
2:  public:  
3:    bool isStrobogrammatic(string num) {  
4:      if (num.size() == 0) return true;  
5:      int l = 0, r = num.size()-1;  
6:      while (l < r) {  
7:        if ((num[l] == '6' && num[r] == '9') || (num[l] == '9' && num[r] == '6') ||  
8:           (num[l] == num[r] && num[l] == '1') || (num[l] == num[r] && num[l] == '8' ||  
9:           (num[l] == num[r] && num[l] == '0'))) {  
10:          l++; r--;     
11:        } else {  
12:          return false;  
13:        }  
14:      }  
15:      if (num.size() & 1) return num[l] == '8' || num[l] == '1' || num[l] == '0';  
16:      return true;  
17:    }  
18:  };  

Then I read the top rated solution and realized that we can set up a look up table first and the code becomes very concise.

1:  class Solution {  
2:  public:  
3:    bool isStrobogrammatic(string num) {  
4:      unordered_map<char, char> dict{{'0', '0'}, {'1', '1'}, {'6', '9'}, {'8', '8'}, {'9', '6'}};  
5:      int n = num.length();  
6:      for (int l = 0, r = n - 1; l <= r; l++, r--)  
7:        if (dict.find(num[l]) == dict.end() || dict[num[l]] != num[r])  
8:          return false;   
9:      return true;   
10:    }  
11:  };