From b73a7da7585b03b42ab51515f5463259ac2e0631 Mon Sep 17 00:00:00 2001
From: Guanzhong Pan <gpan20020205@gmail.com>
Date: Fri, 15 Apr 2022 18:25:41 +0100
Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=201005.K=E6=AC=A1=E5=8F=96?=
 =?UTF-8?q?=E5=8F=8D=E5=90=8E=E6=9C=80=E5=A4=A7=E5=8C=96=E7=9A=84=E6=95=B0?=
 =?UTF-8?q?=E7=BB=84=E5=92=8C.md=20C=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 ...1005.K次取反后最大化的数组和.md | 39 +++++++++++++++++++
 1 file changed, 39 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 45f186e2..0d122d91 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
@@ -211,5 +211,44 @@ var largestSumAfterKNegations = function(nums, k) {
 };
 ```
 
+### C
+```c
+#define abs(a) (((a) > 0) ? (a) : (-(a)))
+
+// 对数组求和
+int sum(int *nums, int numsSize) {
+    int sum = 0;
+
+    int i;
+    for(i = 0; i < numsSize; ++i) {
+        sum += nums[i];
+    }
+    return sum;
+}
+
+int cmp(const void* v1, const void* v2) {
+    return abs(*(int*)v2) - abs(*(int*)v1);
+}
+
+int largestSumAfterKNegations(int* nums, int numsSize, int k){
+    qsort(nums, numsSize, sizeof(int), cmp);
+
+    int i;
+    for(i = 0; i < numsSize; ++i) {
+        // 遍历数组，若当前元素<0则将当前元素转变，k--
+        if(nums[i] < 0 && k > 0) {
+            nums[i] *= -1;
+            --k;
+        }
+    }
+
+    // 若遍历完数组后k还有剩余（此时所有元素应均为正），则将绝对值最小的元素nums[numsSize - 1]变为负
+    if(k % 2 == 1)
+        nums[numsSize - 1] *= -1;
+
+    return sum(nums, numsSize);
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>