From 03b37fee5e16163eb9ff17a9887764d6ba9665c9 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=A4=A7=E7=9C=BC=E7=9D=9B=E5=9B=BE=E5=9B=BE?=
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Date: Tue, 25 Oct 2022 09:22:21 +0800
Subject: [PATCH] =?UTF-8?q?Update=201005.K=E6=AC=A1=E5=8F=96=E5=8F=8D?=
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js 版本，更容易理解，并且用了js es6的语法进行了微调
---
 ...1005.K次取反后最大化的数组和.md | 24 +++++++++----------
 1 file changed, 12 insertions(+), 12 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 9d82bf9f..ed5d1509 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
@@ -190,23 +190,23 @@ func largestSumAfterKNegations(nums []int, K int) int {
 ### Javascript 
 ```Javascript
 var largestSumAfterKNegations = function(nums, k) {
-    nums.sort((a, b) => {
-        return Math.abs(b) - Math.abs(a)
-    })
-    for(let i = 0; i < nums.length; i++) {
-        if(nums[i] < 0 && k > 0) {
-            nums[i] *= -1
-            k--
+
+    nums.sort((a,b) => Math.abs(b) - Math.abs(a))
+    
+    for(let i = 0 ;i < nums.length; i++){
+        if(nums[i] < 0 && k > 0){
+            nums[i] = - nums[i];
+            k--;
         }
     }
 
-    if(k > 0 && k % 2 === 1) {
-        nums[nums.length - 1] *= -1   
+    // 若k还大于0,则寻找最小的数进行不断取反
+    while( k > 0 ){
+        nums[nums.length-1] = - nums[nums.length-1]
+        k--;
     }
-    k = 0
-
     return nums.reduce((a, b) => {
-        return a + b
+         a + b
     })
 };