[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[/i], generate the n[/i]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 ‘’. 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:    }