567. 字符串的排列
给你两个字符串 s1 和 s2 ,写一个函数来判断 s2 是否包含 s1 的排列。如果是,返回 true ;否则,返回 false 。
换句话说,s1 的排列之一是 s2 的 子串 。
示例 1:
输入:s1 = “ab” s2 = “eidbaooo”
输出:true
解释:s2 包含 s1 的排列之一 (“ba”).
示例 2:
输入:s1= “ab” s2 = “eidboaoo”
输出:false
提示:
1 <= s1.length, s2.length <= 104
s1 和 s2 仅包含小写字母
代码:
from leetcode_python.utils import *
class Solution:
def __init__(self):
pass
def checkInclusion(self, s1: str, s2: str) -> bool:
import collections
counter_s1 = collections.Counter(s1)
len_s1 = len(s1)
return any([counter_s1 == collections.Counter(s2[i: i + len_s1]) for i in range(0, len(s2) - len_s1 + 1)])
def test(data_test):
s = Solution()
return s.checkInclusion(*data_test)
def test_obj(data_test):
result = [None]
obj = Solution(*data_test[1][0])
for fun, data in zip(data_test[0][1::], data_test[1][1::]):
if data:
res = obj.__getattribute__(fun)(*data)
else:
res = obj.__getattribute__(fun)()
result.append(res)
return result
if __name__ == '__main__':
datas = [
["ab","eidbaooo"],
]
for data_test in datas:
t0 = time.time()
print('-' * 50)
print('input:', data_test)
print('output:', test(data_test))
print(f'use time:{time.time() - t0}s')
备注:
GitHub:https://github.com/monijuan/leetcode_python
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