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[1/1] 290. Word Pattern
본문
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 <= 300patterncontains only lower-case English letters.1 <= s.length <= 3000scontains only lowercase English letters and spaces' '.<li style="border: 0px solid; box-sizing: border-box; --tw-border-spacing-x:0; --tw-border-spacing-y:0; --tw-translate-x:0; --tw-translate-y:0; --tw-rotate:0; --tw-skew-x:0; --tw-skew-y:0; --tw-scale-x:1; --tw-scale-y:1; --tw-pan-x: ; --tw-pan-y: ; --tw-pinch-zoom: ; --tw-scroll-snap-strictness:proximity; --tw-ordinal: ; --tw-slashed-zero: ;
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class Solution:
def wordPattern(self, pattern: str, s: str) -> bool:
pat_to_word = {}
slist = s.split()
word_to_pat = {}
if len(pattern) != len(slist): return False
for i in range(len(slist)):
if pattern[i] not in pat_to_word:
pat_to_word[pattern[i]] = slist[i]
elif pat_to_word[pattern[i]] != slist[i]:
return False
if slist[i] not in word_to_pat:
word_to_pat[slist[i]] = pattern[i]
elif word_to_pat[slist[i]] != pattern[i]:
return False
return True
