Count and Say leetcode

题目链接

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 nth sequence.

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

题意是n=1时输出字符串1；n=2时，数上次字符串中的数值个数，因为上次字符串有1个1，所以输出11；n=3时，由于上次字符是11，有2个1，所以输出21；n=4时，由于上次字符串是21，有1个2和1个1，所以输出1211。依次类推，写个countAndSay(n)函数返回字符串。

http://blog.csdn.net/kenden23/article/details/17081853

https://github.com/krystism/leetcode/tree/master/algorithms/CountandSay

http://www.cnblogs.com/huxiao-tee/p/4109596.html

注意：用n(int)+'0'表示'n'，仅当n<10时有效！

class Solution {
public:
    string countAndSay(int n) {
        if(==n)
            return NULL;
        string result="";
        while(--n){
            result=nextSequence(result);
        }
        return result;
    }
private:
    string  nextSequence(string s1){
        char cur=s1[];
        int count=;
        string result;
        for(int i=;i<s1.size();i++){
            if(cur!=s1[i]){
                result+=itos(count);
                result.push_back(cur);
                cur=s1[i];
                count=;
            }
            else{
                count++;
            }
        }
        result+=itos(count);
        result.push_back(cur);
        return result;
    }
    string itos(int i){
        ss.str("");
        ss.clear();
        ss<<i;
        return ss.str();
    }
    stringstream ss;
};