Leetcode题目：Count and Say

题目：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 n^th sequence.

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

题目解答：从这个题目中可以很简单的看出，第n个序列得到的字符串与第(n-1)个字符串之间的关系密不可分。为了得到第n个序列，需要从第一个开始向后依次迭代至n。另外需要做的一件事情，就是统计相同的字符出现了几次，为此，设置一个临时变量count。

代码如下：

class Solution {
public:
    string countAndSay(int n) {
        string res = "1";
        if(n <= 0)
            return "";
        else
        {
            convertToString(res,n);
            return res;
        }
    }
   
    void convertToString(string &res, int n)
    {
        for(int i = 1;i < n;i++)
        {
            stringstream res_tmp;
            res_tmp.clear();
            int len = res.length();
            char last = res[0];
            int j = 1;
            int count = 1;
            while(true)
            {
                while((j < len) && (res[j] == last))
                {
                    count++;
                    j++;
                }
                res_tmp << count;
                res_tmp << last;
                if(j < len)
                    last = res[j];
                else
                    break;
                count = 0;
            }
            res = res_tmp.str();
           
        }
    }
};

注意初始时将count置为1，后续设置其值时，设置它为0。

Leetcode题目：Count and Say

原文：http://www.cnblogs.com/CodingGirl121/p/5413934.html