Re: [閒聊] 每日leetcode

[dont (dont)]

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
```

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