From 51d8be645aab9f15101c759535785492c4162e58 Mon Sep 17 00:00:00 2001
From: Zeeland <287017217@qq.com>
Date: Sat, 5 Nov 2022 18:08:43 +0800
Subject: [PATCH] =?UTF-8?q?Update=200242.=E6=9C=89=E6=95=88=E7=9A=84?=
 =?UTF-8?q?=E5=AD=97=E6=AF=8D=E5=BC=82=E4=BD=8D=E8=AF=8D.md?=
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删除一行不必要的print，简化for循环代码
---
 problems/0242.有效的字母异位词.md | 9 ++++-----
 1 file changed, 4 insertions(+), 5 deletions(-)

diff --git a/problems/0242.有效的字母异位词.md b/problems/0242.有效的字母异位词.md
index f8ebc545..b7100b52 100644
--- a/problems/0242.有效的字母异位词.md
+++ b/problems/0242.有效的字母异位词.md
@@ -123,12 +123,11 @@ Python：
 class Solution:
     def isAnagram(self, s: str, t: str) -> bool:
         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
-            record[ord(s[i]) - ord("a")] += 1
-        print(record)
-        for i in range(len(t)):
-            record[ord(t[i]) - ord("a")] -= 1
+            record[ord(i) - ord("a")] += 1
+        for i in t:
+            record[ord(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。