Leetcode 567. Permutation in String
文章作者:Tyan
博客:noahsnail.com | CSDN | 简书
1. Description
2. Solution
**解析:**Version 1,此题与leetcode 438非常类似,思路是一样的。判断s2
是否包含s1
的变换,可以采用字典的方法,即每个字母的个数及类型相等。先统计字符串s1
的字母个数并记录其长度在stat
中,遍历字符串s2
,如果字母在stat
中,则将其记录到字典subs
中,否则重置subs
,当subs['length'] = stat['length']
时,比较二者是否相等,如果相等,直接返回True
,否则,字符串继续遍历,为保证subs
长度与stat
长度一致,此时,subs
中移除s2[index - n + 1]
字符,同时长度减1
。
Version 1
class Solution:
def checkInclusion(self, s1: str, s2: str) -> bool:
n = len(s1)
stat = collections.Counter(s1)
stat['length'] = n
subs = collections.defaultdict(int)
for index, ch in enumerate(s2):
if ch in stat:
subs[ch] += 1
subs['length'] += 1
else:
subs = collections.defaultdict(int)
continue
if subs['length'] == stat['length']:
if stat == subs:
return True
subs[s2[index - n + 1]] -= 1
subs['length'] -= 1
return False
Reference
- https://leetcode.com/problems/permutation-in-string/
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