Q38 Count and Say

1.   1
2.   11
3.   21
4.   1211
5.   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.

Input: 1
Output: "1"

Input: 4
Output: "1211"

class Solution:
    def countAndSay(self, n):
        """
        :type n: int
        :rtype: str
        """
        if n == 1:
            return '1'
        i = 1          # 从1开始构造
        lastStr = '1'  # 上一个字符串随着迭代次数更新
        while n > i:
            temp = ''  # 临时子串
            count = 0  # 统计相同字符的个数
            for j in range(len(lastStr)):  # 找出上一个字符串的特点，如 5 -> 111221
                if j > 0 and lastStr[j] != lastStr[j-1]:
                    temp += str(count) + lastStr[j-1] 
                    count = 1  # 重置count
                else:
                    count += 1              
            lastStr = temp + str(count) + lastStr[j]  # 构造当前字符串，如 6 -> 312211
            i += 1
        return lastStr

a = 7
b = Solution()
print(b.countAndSay(a))  # 13112221，因为6对应的字符串为312211