[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<) return NULL;

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