66. Plus One

I was thinking of add 1 from end and insert 1 in the front if any carry. But I found the top rated solution is very concise and clever. Pushing a number to the back is much faster than inserting in the front.

1:  class Solution {  
2:  public:  
3:    vector<int> plusOne(vector<int>& digits) {  
4:      int n = digits.size();  
5:      for (int i = n-1; i >= 0; i--) {  
6:        if (digits[i] == 9) digits[i] = 0;  
7:        else { digits[i]++; return digits; }  
8:      }  
9:      digits[0] = 1;  
10:      digits.push_back(0);  
11:      return digits;  
12:    }  
13:  };