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Update 0242.有效的字母异位词.md

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删除一行不必要的print，简化for循环代码
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Zeeland
2022-11-05 18:08:43 +08:00
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parent fe16fac72f
commit 51d8be645a
1 changed files with 4 additions and 5 deletions
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problems/0242.有效的字母异位词.md
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@ -123,12 +123,11 @@ Python：
class Solution: class Solution:
def isAnagram(self, s: str, t: str) -> bool: def isAnagram(self, s: str, t: str) -> bool:
record = [0] * 26 record = [0] * 26
for i in range(len(s)): for i in 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 #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 record[ord(i) - ord("a")] += 1
print(record) for i in t:
for i in range(len(t)): record[ord(i) - ord("a")] -= 1
record[ord(t[i]) - ord("a")] -= 1
for i in range(26): for i in range(26):
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。