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Merge pull request #1736 from Undertone0809/Undertone0809-patch-1
Update 0242.有效的字母异位词.md
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@ -123,12 +123,11 @@ 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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for i in 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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record[ord(i) - ord("a")] += 1
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for i in t:
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record[ord(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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