Leetcode 38 Count and Say

题目要求：

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由于n=1的结果为1，有1个1，所以返回11

n=3由于n=2结果为11，有2个1，返回21

n=4由于n=3结果为21，有1个2和1个1，所以返回1211

给定n,以此类推

解答：遍历每个子串，根据统计个数（如1115，计算为3115）衍生出新的子串，循环遍历，返回最后一次的结果

代码如下：

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;
                }
                //统计每个数字的个数
                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;
    }
};