Count and Say @python 题解

题意描述：

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.

题意就是迭代生成字符串：后一个字符串是前一个字符串每个连续出现的数字的个数＋该数字拼凑而成，用count记录连续数字的个数，然后用key记录当前最新的数字，每次遇到不同的数字就把count+key拼接到结果当中，但注意如果遍历到最后一位时，无论是否遇到新的字符都要把当前count+key加入结果中
代码如下

class Solution:    # @param n, an integer    # @return a string    def countAndSay(self, n):        if n==1: return '1'        beforeSeq='1';afterSeq=''        while n>1:            curLength=len(beforeSeq)            key=beforeSeq[0];count=0            afterSeq=''            for i in range(curLength):                if beforeSeq[i]==key:                    count+=1                    if i==curLength-1:                        afterSeq+=(str(count)+key)                else:                    afterSeq+=(str(count)+key)                    key=beforeSeq[i]                    count=1                    if i==curLength-1:                        afterSeq+=(str(count)+key)            beforeSeq=afterSeq            n-=1        return afterSeq