【LeetCode with Python】 Count and Say

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原题页面：https://oj.leetcode.com/problems/count-and-say/
题目类型：字符串
难度评价：★
本文地址：http://blog.csdn.net/nerv3x3/article/details/39453287

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 n^th sequence.

Note: The sequence of integers will be represented as a string.

主要是计数的逻辑。每一轮，遇到一个不同的字符时，就结束到该轮的计数。但是需要注意的是最后一轮的计数，因为遇到的是字符串src的结尾，所以需要强制结束这一轮的计数，就类似于编译原理词法分析最后的那个$符号的作用。

class Solution:    def doCountAndSay(self, src):        char = src[0]        num = 0        result = ""        for c in src:            if char == c:                num += 1            else:                result += (str(num) + char)                char = c                num = 1        result += (str(num) + char)        return result    # @return a string    def countAndSay(self, n):        if 0 == n:            return ""        elif 1 == n:            return "1"        result = "1"         # str is a keyword        for i in range(1, n):            result = self.doCountAndSay(result)        return result