[Leetcode] count and say 计数和说

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

1is read off as"one 1"or11.
11is read off as"two 1s"or21.
21is read off as"one 2, thenone 1"or1211.

Given an integer n, generate the n th sequence.

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

题意：返回第n个序列，第i+1个字符串是第i个字符串的读法。参考：Grandyang和JustDoIT的博客。

思路：算法就是对于前一个数，找出相同元素的个数，把个数和该元素存到新的string里。代码需要两个循环，第一个是为找到第n个，第二是为了，根据上一字符串的信息来实现当前的字符串。

class Solution {
public:
    string countAndSay(int n) 
    {
        if(n<1) return NULL;

        string res="1";
        for(int i=1;i<n;++i)
        {
            string temp;    //当前序列
            res.push_back('*');
            int count=0;    //重复的个数
            for(int j=0;j<res.size();++j)
            {
                if(j==0)
                    count++;
                else
                {
                    if(res[j] !=res[j-1])
                    {
                        temp.push_back(count+'0');
                        temp.push_back(res[j-1]);
                        count=1;
                    }
                    else
                        ++count;
                }    
            }
            res=temp;  
        }
        return res;
    }
};