报数

Count and Say

报数序列是指一个整数序列，按照其中的整数的顺序进行报数，得到下一个数。其前五项如下：

	1.     1
	2.     11
	3.     21
	4.     1211
	5.     111221

1 被读作 "one 1" ("一个一") , 即 11。

11 被读作 "two 1s" ("两个一"）, 即 21。

21 被读作 "one 2", "one 1" （"一个二" , "一个一") , 即 1211。

给定一个正整数 n ，输出报数序列的第 n 项。

注意：整数顺序将表示为一个字符串。

英文题目:

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

Input: 6

Output: "312211"

思路

say变量保存第n-1个报数序列
从遍历say字符串，say[0]保存为curr_num，用count计数，当遇到不为curr_num的元素x时，得到报数序列str(count) + curr_num
把count置为1，因为遇到不为curr_num的元素x时，x已经出现一次，同时curr_num = x
从n=2开始循环上述过程到n

代码

class Solution:
    def countAndSay(self, n):
        """
        :type n: int
        :rtype: str
        """
        say = '1'
        tmp = ''
        for _ in range(1, n):
            curr_num = say[0]
            count = 0
            for x in say:
                if x == curr_num:
                    count += 1 #count
                else:
                    tmp = tmp + str(count) + curr_num #say
                    curr_num = x
                    count = 1 # x != curr_num时, x已经出现过一次
            tmp = tmp + str(count) + curr_num
            say, tmp = tmp, ''
        return say

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