-
Notifications
You must be signed in to change notification settings - Fork 0
/
38.报数.py
76 lines (72 loc) · 1.62 KB
/
38.报数.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
#
# @lc app=leetcode.cn id=38 lang=python3
#
# [38] 报数
#
# https://leetcode-cn.com/problems/count-and-say/description/
#
# algorithms
# Easy (48.39%)
# Total Accepted: 22.9K
# Total Submissions: 47.3K
# Testcase Example: '1'
#
# 报数序列是一个整数序列,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(1 ≤ n ≤ 30),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。
#
#
#
# 示例 1:
#
# 输入: 1
# 输出: "1"
#
#
# 示例 2:
#
# 输入: 4
# 输出: "1211"
#
#
#
class Solution:
def tongji(self, zifuchuan):
mylen = len(zifuchuan)
jieguo = []
count = 1
for i in range(mylen-1):
if zifuchuan[i] == zifuchuan[i+1]:
count += 1
else:
jieguo.append(str(count))
jieguo.append(zifuchuan[i])
count = 1
jieguo.append(str(count))
jieguo.append(zifuchuan[mylen-1])
return jieguo
def countAndSay(self, n: int) -> str:
if n <= 1:
return "1"
if n == 2:
return "11"
temlist=['1','1']
for i in range(3,n+1):
temlist=self.tongji(temlist)
res=""
for temm in temlist:
res=res.__add__(temm)
return res