From fac95a0a430ae42ea682fc6a89dab9b2394230f1 Mon Sep 17 00:00:00 2001
From: Violet_Fu <156555169+iamziqian@users.noreply.github.com>
Date: Mon, 29 Jul 2024 22:07:20 -0700
Subject: [PATCH 1/2] =?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

Java的版本二题解
---
 ...1005.K次取反后最大化的数组和.md | 30 +++++++++++++++++++
 1 file changed, 30 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 fa27d3b7..4e049600 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
@@ -128,6 +128,36 @@ class Solution {
 
     }
 }
+
+// 版本二：排序数组并贪心地尽可能将负数翻转为正数，再根据剩余的k值调整最小元素的符号，从而最大化数组的总和。
+class Solution {
+    public int largestSumAfterKNegations(int[] nums, int k) {
+        if (nums.length == 1) return nums[0];
+
+        // 排序：先把负数处理了
+        Arrays.sort(nums); 
+
+        for (int i = 0; i < nums.length && k > 0; i++) { // 贪心点, 通过负转正, 消耗尽可能多的k
+            if (nums[i] < 0) {
+                nums[i] = -nums[i];
+                k--;
+            }
+        }
+
+        // 退出循环, k > 0 || k < 0 (k消耗完了不用讨论)
+        if (k % 2 == 1) { // k > 0 && k is odd：对于负数：负-正-负-正
+            Arrays.sort(nums); // 再次排序得到剩余的负数，或者最小的正数
+            nums[0] = -nums[0];
+        }
+        // k > 0 && k is even，flip数字不会产生影响: 对于负数: 负-正-负；对于正数：正-负-正 
+
+        int sum = 0;
+        for (int num : nums) { // 计算最大和
+            sum += num;
+        }
+        return sum;
+    }
+}
 ```
 
 ### Python

From c49dd3b1595fe4b8e2adb991b0882a78774ca32f Mon Sep 17 00:00:00 2001
From: Violet_Fu <156555169+iamziqian@users.noreply.github.com>
Date: Tue, 30 Jul 2024 13:21:10 -0700
Subject: [PATCH 2/2] =?UTF-8?q?Update=200134.=E5=8A=A0=E6=B2=B9=E7=AB=99.m?=
 =?UTF-8?q?d?=
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0134.加油站的Java解法三
---
 problems/0134.加油站.md | 23 +++++++++++++++++++++++
 1 file changed, 23 insertions(+)

diff --git a/problems/0134.加油站.md b/problems/0134.加油站.md
index c88b43b1..1329bd9a 100644
--- a/problems/0134.加油站.md
+++ b/problems/0134.加油站.md
@@ -249,6 +249,29 @@ class Solution {
     }
 }
 ```
+```
+// 解法3
+class Solution {
+    public int canCompleteCircuit(int[] gas, int[] cost) {
+        int tank = 0; // 当前油量
+        int totalGas = 0;  // 总加油量
+        int totalCost = 0; // 总油耗
+        int start = 0; // 起点
+        for (int i = 0; i < gas.length; i++) {
+            totalGas += gas[i];
+            totalCost += cost[i];
+            
+            tank += gas[i] - cost[i];
+            if (tank < 0) { // tank 变为负数 意味着 从0到i之间出发都不能顺利环路一周，因为在此i点必会没油
+                tank = 0; // reset tank，类似于题目53.最大子树和reset sum
+                start = i + 1; // 起点变为i点往后一位
+            }
+        }
+        if (totalCost > totalGas) return -1;
+        return start;
+    }
+}
+```
 
 ### Python 
 暴力法