题目地址：https://leetcode.com/problems/license-key-formatting/description/open in new window
题目描述
Youare given a license key represented as a string S which consists only alphanumeric character and dashes. The string is separated into N+1 groups by N dashes.
Given a number K, we would want to reformat the strings such that each group contains exactly K characters, except for the first group which could be shorter than K, but still must contain at least one character. Furthermore, there must be a dash inserted between two groups and all lowercase letters should be converted to uppercase.
Given a non-empty string S and a number K, format the string according to the rules described above.
Example 1:
Input: S = "5F3Z-2e-9-w", K = 4
Output: "5F3Z-2E9W"
Explanation: The string S has been split into two parts, each part has 4 characters.
Note that the two extra dashes are not needed and can be removed.
Example 2:
Input: S = "2-5g-3-J", K = 2
Output: "2-5G-3J"
Explanation: The string S has been split into three parts, each part has 2 characters except the first part as it could be shorter as mentioned above.
Note:
1、 ThelengthofstringSwillnotexceed12,000,andKisapositiveinteger.；
2、 StringSconsistsonlyofalphanumericalcharacters(a-zand/orA-Zand/or0-9)anddashes(-).；
3、 StringSisnon-empty.；
题目大意
现在有一些用-分割的字符串，需要重新安排，使得除了第一个之外，其他的-分割的字符串长度都是K。另外需要全部转成大写字符。
解题方法
注意，这个题的意思是 右边的序列要都是K个的，最左边如果不够就不够了，剩多少写多少。
首先计算第一个应该占据了多少个字符，然后看后面的应该是等长的。并且和原来的-的划分情况是无关的。每个片的个数都要是K。 字符串切片结束的长度大于自身长度也可以。
class Solution(object):
def licenseKeyFormatting(self, S, K):
"""
:type S: str
:type K: int
:rtype: str
"""
S = S.upper()
groups = ''.join(S.split('-'))
bias = len(groups) % K
devides = len(groups) / K
answer = groups[:bias]
answer += '-' if bias != 0 else ''
for i in range(devides):
answer += groups[i*K+bias : (i+1)*K+bias] + '-'
return answer[:-1]
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
二刷的时候，写的Python代码如下：
class Solution(object):
def licenseKeyFormatting(self, S, K):
"""
:type S: str
:type K: int
:rtype: str
"""
res = []
s = "".join(S.split("-")).upper()
N = len(s)
if N % K != 0:
res.append(s[: N % K])
for i in range(N % K, N, K):
res.append(s[i : i + K])
return "-".join(res)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
DDKK.COM 弟弟快看-教程，程序员编程资料站，版权归原作者所有
本文经作者：负雪明烛 授权发布，任何组织或个人未经作者授权不得转发