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.

題意：返回第n個序列，第i+1個字符串是第i個字符串的讀法。參考：Grandyang和JustDoIT的博客。

思路：算法就是對於前一個數，找出相同元素的個數，把個數和該元素存到新的string裏。代碼需要兩個循環，第一個是為找到第n個，第二是為了，根據上一字符串的信息來實現當前的字符串。

 1 class Solution {
 2 public:
 3     string countAndSay(int n) 
 4     {
 5         if(n<1) return NULL;
 6 
 7         string res="1";
 8         for
 
(int i=1;i<n;++i)
 9         {
10             string temp;    //當前序列
11             res.push_back(‘*‘);
12             int count=0;    //重復的個數
13             for(int j=0;j<res.size();++i)
14             {
15                 if(j==0)
16                     count++;
17                 else
18                 {
 
19                     if(res[j] !=res[j-1])
20                     {
21                         temp.push_back(count+‘0‘);
22                         temp.push_back(res[j-1]);
23                         count=1;
24                     }
25                     else
26                         ++count;
27                 }    
28             }
29             res=temp;  
30         }
31         return res;
32     }
33 };

[Leetcode] count and say 計數和說