From 440bddb44bc90d0a42b0834903eb77bfaa565377 Mon Sep 17 00:00:00 2001
From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com>
Date: Thu, 1 Jun 2023 00:45:16 -0500
Subject: [PATCH] =?UTF-8?q?Update=201005.K=E6=AC=A1=E5=8F=96=E5=8F=8D?=
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---
 .../1005.K次取反后最大化的数组和.md | 18 ++++++++++++------
 1 file changed, 12 insertions(+), 6 deletions(-)

diff --git a/problems/1005.K次取反后最大化的数组和.md b/problems/1005.K次取反后最大化的数组和.md
index 27a575c7..343bef04 100644
--- a/problems/1005.K次取反后最大化的数组和.md
+++ b/problems/1005.K次取反后最大化的数组和.md
@@ -155,17 +155,23 @@ class Solution {
 ```
 
 ### Python
+贪心
 ```python
 class Solution:
     def largestSumAfterKNegations(self, A: List[int], K: int) -> int:
-        A = sorted(A, key=abs, reverse=True) # 将A按绝对值从大到小排列
-        for i in range(len(A)):
-            if K > 0 and A[i] < 0:
+        A.sort(key=lambda x: abs(x), reverse=True)  # 第一步：按照绝对值降序排序数组A
+
+        for i in range(len(A)):  # 第二步：执行K次取反操作
+            if A[i] < 0 and K > 0:
                 A[i] *= -1
                 K -= 1
-        if K > 0:
-            A[-1] *= (-1)**K #取A最后一个数只需要写-1
-        return sum(A)
+
+        if K % 2 == 1:  # 第三步：如果K还有剩余次数，将绝对值最小的元素取反
+            A[-1] *= -1
+
+        result = sum(A)  # 第四步：计算数组A的元素和
+        return result
+
 ```
 
 ### Go