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 nth sequence.

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

分析

这是一道根据规则推导题目，要求给定序列数n，求出该序列对应的字符串。

规则如上图所示。

AC代码

class Solution {
public:
    string countAndSay(int n) {
        if (n <= 0)
            return NULL;

        //n=1时，结果为"1"
        string ret = "1";
        if (n == 1)
            return ret;
        else
        {
            for (int i = 2; i <= n; i++)
                ret = Count(ret);
        }//else
        return ret;

    }

    string Count(const string &str)
    {
        int size = strlen(str.c_str());
        //保存结果
        stringstream ret;
        //保存标识字符
        char flag = str[0];
        //计算标识字符的出现次数
        int count = 0 , i = 0;
        while( i < size )
        {
            //临时循环位
            int pos = i;
            while (str[pos] == flag)
            {
                count++;
                pos++;
            }//while
            ret << count << flag;
            flag = str[pos];
            count = 0;
            //设置下一个循环位
            i = pos;
        }//for
        return ret.str();
    }
};

GitHub测试程序源码

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