From 175f1c5e1a459e72511ae56ce33828a8fc367690 Mon Sep 17 00:00:00 2001
From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com>
Date: Mon, 16 Aug 2021 19:11:16 +0800
Subject: [PATCH] =?UTF-8?q?=E7=AE=80=E5=8C=96=200001.=E4=B8=A4=E6=95=B0?=
 =?UTF-8?q?=E4=B9=8B=E5=92=8C.md=20python=E4=BB=A3=E7=A0=81?=
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原代码过于赘余，需要遍历两次且可读性较差。更新后的代码在维持题主题意的基础上的一次优化
---
 problems/0001.两数之和.md | 15 ++++++++-------
 1 file changed, 8 insertions(+), 7 deletions(-)

diff --git a/problems/0001.两数之和.md b/problems/0001.两数之和.md
index f8c9da5f..9b961d0b 100644
--- a/problems/0001.两数之和.md
+++ b/problems/0001.两数之和.md
@@ -110,13 +110,14 @@ Python：
 ```python
 class Solution:
     def twoSum(self, nums: List[int], target: int) -> List[int]:
-        hashmap={}
-        for ind,num in enumerate(nums):
-            hashmap[num] = ind
-        for i,num in enumerate(nums):
-            j = hashmap.get(target - num)
-            if j is not None and i!=j:
-                return [i,j]
+        records = dict()
+
+        # 用枚举更方便，就不需要通过索引再去取当前位置的值
+        for idx, val in enumerate(nums):
+            if target - val not in records:
+                records[val] = idx
+            else:
+                return [records[target - val], idx] # 如果存在就返回字典记录索引和当前索引
 ```