Re: [閒聊] 每日leetcode
1813. Sentence Similarity III
## 思路
Case1. prefix == s1, 在後面加字串 (aa, aabcd)
Case2. suffix == s1, 在前面加字串 (aa, bcdaa)
Case3. prefix+suffix == s1, 在中間加字串 (aa, abcda)
先把兩個字串都轉成word陣列
用two pointer檢查s1跟s2的prefix跟suffix相同words個數
如果超過s1的word個數就回傳True
## Code
```python
class Solution:
def areSentencesSimilar(self, sentence1: str, sentence2: str) ->
bool:
# aa aabcd True
# aa bcdaa True
# aa baacd False
# aa abcda True
words1 = sentence1.split()
words2 = sentence2.split()
if len(words1) > len(words2):
words1, words2 = words2, words1
left, right = 0, len(words1) - 1
count = len(words1)
for i in range(len(words2)):
if words2[i] != words1[left]:
break
left += 1
count -= 1
if count == 0:
return True
for i in range(len(words2)-1, -1, -1):
if words2[i] != words1[right]:
break
right -= 1
count -= 1
if count == 0:
return True
return False
```
--
https://i.imgur.com/kyBhy6o.jpeg
--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 185.213.82.123 (臺灣)
※ 文章網址: https://www.ptt.cc/bbs/Marginalman/M.1728178520.A.684.html
討論串 (同標題文章)
完整討論串 (本文為第 950 之 1554 篇)：