leetcode count and say

问题描述：

https://oj.leetcode.com/problems/count-and-say/ 点击打开链接

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.

问题分析：

通过最当前数字序列cur数数, 产生下一个序列next;

1(1个1) --> 11(2个1)--> 21-->  1211

对于输入n, 需要迭代 n - 1 次得到所需要的序列，序列产生的方法

如上边所示。重点在于统计字符s[i]出现的次数，重新拼接成新的字符串。

示例代码：

string countAndSay(int n)     {        string cur = "1";        for (int i = 1; i < n; i++)        {               int m = cur.length();            int num = 1;            char ch = cur[0];            stringstream next; /* 采用字符串流进行辅助字符串拼组 */            for (int j = 1; j < m; j++)            {                if (cur[j] == ch)                {   num++;  } /* 统计计数 */                else                {                    next << num << ch; /* 遇到新字符，旧字符统计完成，开始拼接 */                    ch = cur[j]; num = 1;                }            }            next << num << ch;            cur = next.str();        }        return cur;    }