Description
Given a pattern and a string s, find if s follows the same pattern.
Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in s.
Example 1:
Input: pattern = "abba", s = "dog cat cat dog"
Output: true
Example 2:
Input: pattern = "abba", s = "dog cat cat fish"
Output: false
Example 3:
Input: pattern = "aaaa", s = "dog cat cat dog"
Output: false
Constraints:
1 <= pattern.length <= 300
pattern contains only lower-case English letters.
1 <= s.length <= 3000
s contains only lowercase English letters and spaces ' '.
s does not contain any leading or trailing spaces.
All the words in s are separated by a single space.
Solution
Use hashmap to map pattern to s文章来源：https://www.toymoban.com/news/detail-492534.html
Time complexity:
o
(
n
)
o(n)
o(n)
Space complexity:
o
(
n
)
o(n)
o(n)文章来源地址https://www.toymoban.com/news/detail-492534.html
Code
class Solution:
def wordPattern(self, pattern: str, s: str) -> bool:
s = s.split(' ')
if len(pattern) != len(s) or len(set(pattern)) != len(set(s)):
return False
p_to_s = {}
for each_p, each_s in zip(pattern, s):
if each_p not in p_to_s:
p_to_s[each_p] = each_s
elif p_to_s[each_p] != each_s:
return False
return True
到了这里，关于leetcode - 290. Word Pattern的文章就介绍完了。如果您还想了解更多内容，请在右上角搜索TOY模板网以前的文章或继续浏览下面的相关文章，希望大家以后多多支持TOY模板网！
