From 82a58bc20249bb248dbbb0d90771db0ea60bf017 Mon Sep 17 00:00:00 2001
From: Jamcy123 <1219502823@qq.com>
Date: Fri, 6 May 2022 15:49:43 +0800
Subject: [PATCH] =?UTF-8?q?1005.K=E6=AC=A1=E5=8F=96=E5=8F=8D=E5=90=8E?=
 =?UTF-8?q?=E6=9C=80=E5=A4=A7=E5=8C=96=E7=9A=84=E6=95=B0=E7=BB=84=E5=92=8C?=
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---
 .../1005.K次取反后最大化的数组和.md   | 16 ++++++++++++++++
 1 file changed, 16 insertions(+)

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 79767deb..202534da 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
@@ -209,6 +209,22 @@ var largestSumAfterKNegations = function(nums, k) {
         return a + b
     })
 };
+
+// 版本二 (优化: 一次遍历)
+var largestSumAfterKNegations = function(nums, k) {
+    nums.sort((a, b) => Math.abs(b) - Math.abs(a)); // 排序
+    let sum = 0;
+    for(let i = 0; i < nums.length; i++) {
+        if(nums[i] < 0 && k-- > 0) { // 负数取反（k 数量足够时）
+            nums[i] = -nums[i];
+        }
+        sum += nums[i]; // 求和
+    }
+    if(k % 2 > 0) { // k 有多余的（k若消耗完则应为 -1）
+        sum -= 2 * nums[nums.length - 1]; // 减去两倍的最小值（因为之前加过一次）
+    }
+    return sum;
+};
 ```