【LeetCode题意分析&解答】38. Count and Say
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.
Note: The sequence of integers will be represented as a string.
题意分析:
本题是将数字从1开始,将当前数字转化为口语对应的数字。比如1口语是1个1,记作11;11读作2个1,记作21;21读作1个2,1个1,记作1211……
'1'是第一个数字,根据输入的数字n,计算第n个这样的数字。
说起来比较拗口,对照着例子可以感受一下……
解答:
如果你理解题意,那么解答就很简单。直接循环,每个数字按照题目要求翻译成口语对应的数字,按照顺序找到第n个返回就行了。
Leetcode给的评级是Easy,这是对于程序难度来说,然而对于题意来说容易理解偏。
AC代码:
class Solution(object):
def countAndSay(self, n):
if n < 2: return ''
ret_str = ''
while n > 1:
temp, current_num = '', 0
for i, v in enumerate(ret_str):
if i > 0 and v != ret_str[i - 1]:
temp += str(current_num) + ret_str[i - 1]
current_num = 1
else:
current_num += 1
ret_str = temp + (str(current_num) + ret_str[-1] if current_num != 0 else '')
n -= 1
return ret_str
后话:
其实这道题有个很有意思的地方,即输出的数字中永远不可能有大于3的数字,只会出现1、2、3三个数字,你能够证明吗?
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