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Merge pull request #1497 from geek-yang/patch-1 

修改: (0242. 有效的字母异位词): 修改python解法
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程序员Carl 2022-07-31 10:24:58 +08:00 committed by GitHub
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problems/0242.有效的字母异位词.md
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@ -127,8 +127,6 @@ class Solution:
if record[i] != 0: 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。 #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 False
#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
break
return True return True
``` ```