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Update 0242.有效的字母异位词.md
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@ -125,6 +125,8 @@ class Solution:
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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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#如果有一个元素不为零,则可以判断字符串s和t不是字母异位词
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break
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return True
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```
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