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Update 0242.有效的字母异位词.md
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@ -113,7 +113,22 @@ class Solution {
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```
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Python:
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```python
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class Solution:
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def isAnagram(self, s: str, t: str) -> bool:
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record = [0] * 26
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for i in range(len(s)):
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#并不需要记住字符a的ASCII,只要求出一个相对数值就可以了
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record[ord(s[i]) - ord("a")] += 1
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print(record)
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for i in range(len(t)):
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record[ord(t[i]) - ord("a")] -= 1
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for i in range(26):
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if record[i] != 0:
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#record数组如果有的元素不为零0,说明字符串s和t 一定是谁多了字符或者谁少了字符。
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return False
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return True
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```
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Go:
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```go
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