From 4defdbba77db660b43543d7660630bcd44dbb47c Mon Sep 17 00:00:00 2001
From: daniel1n <54945782+daniel1n@users.noreply.github.com>
Date: Sat, 17 Jul 2021 22:45:14 +0800
Subject: [PATCH] =?UTF-8?q?Update=201005.K=E6=AC=A1=E5=8F=96=E5=8F=8D?=
 =?UTF-8?q?=E5=90=8E=E6=9C=80=E5=A4=A7=E5=8C=96=E7=9A=84=E6=95=B0=E7=BB=84?=
 =?UTF-8?q?=E5=92=8C.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

与C++相同的思路，不过排序的耗时更长
---
 ...1005.K次取反后最大化的数组和.md | 27 +++++++++++++++++++
 1 file changed, 27 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 c3e99f7e..3534e1e2 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
@@ -99,6 +99,33 @@ public:
 
 
 Java：
+```java
+class Solution {
+    public int largestSumAfterKNegations(int[] nums, int K) {
+    	// 将数组按照绝对值大小从大到小排序，注意要按照绝对值的大小
+	nums = IntStream.of(nums)
+		     .boxed()
+		     .sorted((o1, o2) -> Math.abs(o2) - Math.abs(o1))
+		     .mapToInt(Integer::intValue).toArray();
+	int len = nums.length;	    
+	for (int i = 0; i < len; i++) {
+	    //从前向后遍历，遇到负数将其变为正数，同时K--
+	    if (nums[i] < 0 && k > 0) {
+	    	nums[i] = -nums[i];
+		k--;
+	    }
+	}
+	// 如果K还大于0，那么反复转变数值最小的元素，将K用完
+	if (k % 2 == 1) nums[len - 1] = -nums[len - 1];
+	int result = 0;
+	for (int a : nums) {
+	    result += a;
+	}
+	return result;
+    }
+}
+```
+
 ```java
 class Solution {
     public int largestSumAfterKNegations(int[] A, int K) {