[LeetCode] Count and Say, Solution

The count-and-say sequence is the sequence of integers beginning as follows:
1, 11, 21, 1211, 111221, ...

1 is read off as "one 1" or 11.
11 is read off as "two 1s" or 21.
21 is read off as "one 2, then one 1" or 1211.

Given an integer n, generate the n^th sequence.

Note: The sequence of integers will be represented as a string.

» Solve this problem

[Thoughts]
string-operation. The only trick thing is Line11. seq[seq.size()] always '\0'. It will help to save an "if" statement.


[Code]

1:    string countAndSay(int n) {  
2:      // Start typing your C/C++ solution below  
3:      // DO NOT write int main() function  
4:      string seq = "1";  
5:      int it = 1;  
6:      while(it<n)  
7:      {  
8:        stringstream newSeq;  
9:        char last = seq[0];  
10:        int count =0;  
11:        for(int i =0; i<= seq.size();i++)  
12:        {  
13:          if(seq[i] ==last)  
14:          {  
15:            count ++;  
16:            continue;  
17:          }  
18:          else  
19:          {  
20:            newSeq<<count<<last;  
21:            last = seq[i];   
22:            count =1;  
23:          }  
24:        }  
25:        seq = newSeq.str();  
26:        it++;  
27:      }  
28:      return seq;  
29:    }