Leetcode 38 Count and Say

題目要求：

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.

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n=1 返回1

n=2由於n=1的結果爲1，有1個1，所以返回11

n=3由於n=2結果爲11，有2個1，返回21

n=4由於n=3結果爲21，有1個2和1個1，所以返回1211

給定n,以此類推

解答：遍歷每個子串，根據統計個數（如1115，計算爲3115）衍生出新的子串，循環遍歷，返回最後一次的結果

代碼如下：

class Solution {
public:
    string countAndSay(int n) {
        string tmp;
        string result;
        char c = 1 + '0';
        result += c;
        tmp += c;
        
        int i = 1;
        while(i < n)
        {
            string tmp1;
            int k = 1;
            for(int j = 0 ;j < tmp.length(); j++)
            {
                if(j+1 == tmp.length())
                {
                    tmp1 += char(k+'0');
                    tmp1 += tmp[j];
                    break;
                }
                //統計每個數字的個數
                if(tmp[j] == tmp[j+1])
                {
                    k++;
                    continue;
                }
                else
                {
                    tmp1 += char(k+'0');
                    tmp1 += tmp[j];
                    k = 1;
                }
            }
            if(tmp1.length() == 0)
            {
                tmp1 += char(k+'0');
                tmp1 += tmp[0];
            }
            i++;
            tmp = tmp1;
            result = tmp1;
        }
        
        return result;
    }
};