Update 0242.有效的字母异位词.md

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Baturu 2021-05-31 18:31:10 -07:00 committed by GitHub
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@ -113,7 +113,22 @@ class Solution {
```
Python
```python
class Solution:
def isAnagram(self, s: str, t: str) -> bool:
record = [0] * 26
for i in range(len(s)):
#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
record[ord(s[i]) - ord("a")] += 1
print(record)
for i in range(len(t)):
record[ord(t[i]) - ord("a")] -= 1
for i in range(26):
if record[i] != 0:
#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。
return False
return True
```
Go
```go