LeetCode小白菜笔记[12]：Count and Say

LeetCode小白菜笔记[12]：Count and Say

38. count and say [easy]

题目如下：

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 *n*th 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"

题目的意思是，对于一个整数序列，每一个数字都是上面的一个count and say。所谓的 count and say 的意思就是说，对于上一个数字，如果有n个连续的数字m，那么下一个数字中就有一个连着的nm表示n个m，然后按顺序依次数完上面的数字。开始以为有规律，后来没有发现规律，放弃，让生成第n个，就需要从第一个开始，做n次这样的处理，得到第n个，code如下：

class Solution(object):
    def countAndSay(self, n):
        """
        :type n: int
        :rtype: str
        """
        if n == 1:
            return '1'
        elif n == 2:
            return '11'
        elif n == 3:
            return '21'
        else:
            last = self.countAndSay(n-1)
            current = ''
            count = 1
            say = last[0]
            for i in range(1,len(last)):
                if last[i] == say:
                    count += 1
                else:
                    current = current + str(count) + str(say)
                    say = last[i]
                    count = 1
            current = current + str(count) + str(say)
            return current

用了个递归，常规解法。为了写range(1,len(last))不报错，把前三项都做成了直接return。最后倒数第二行是因为最后的一组count and say只是数出来了，并未加到返回值里面。因为我们用后一项和前一项不相同作为分割符，然后扩充要返回的字符串，因此最后一组没有分隔符，因为后面没有了，所以要单独加上。

class Solution(object):
    def countAndSay(self, n):
        """
        :type n: int
        :rtype: str
        """
        if n == 1:
            return '1'
        else:
            last = self.countAndSay(n-1)
            current = ''
            count = 1
            say = 0
            for i in range(len(last)):
                if last[i] == say:
                    count += 1
                else:
                    if not say == 0:
                        current = current + str(count) + str(say)
                    say = last[i]
                    count = 1
            current = current + str(count) + str(say)
            return current

上面的elif语句太多，不好看，所以改了一下，这样就只需要一个直接return的，其余都通过递归。

2018年2月8日23:02:09

放假回家第一天~加油加油~