leetcode#38. Count and Say

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The count-and-say sequence is the sequence of integers with the first five terms as following:

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.
Given an integer n, generate the nth term of the count-and-say sequence.

Note: Each term of the sequence of integers will be represented as a string.
Example 1:

Input: 1
Output: "1"

Example 2:

Input: 4
Output: "1211"

这题刚开始没看懂，不知道在搞啥，最后看明白了“数数”呗。从1开始数，1读成一个一，即11；11读成两个一，即21；21读成一个二一个一，即1211，etc。
这题反正我是没看出啥好解法，只能从最开始数了，有人有好的想法的话欢迎留言～

代码

class Solution(object):def countAndSay(self, n):""":type n: int:rtype: str"""tmp = "1"for i in range(1, n):tmp = self.count(tmp)return tmpdef count(self, tmp):flag = tmp[0]cnt = 0ans = ""for i in range(len(tmp)):if flag == tmp[i]:cnt = cnt + 1if i == len(tmp) - 1:ans = ans + str(cnt) + flagelse:ans = ans + str(cnt) + flagflag = tmp[i]cnt = 1if i == len(tmp) - 1:ans = ans + "1" + flagreturn ans

似乎没啥好说的，就是标记，计数然后输出。需要注意的就是最后一个字符，可能和之前的字符不一样，处理时需要单独判断下最后一个字符。

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