添加 problem 1005

Algorithms/1005. Maximize Sum Of Array After K Negations/1005. Maximize Sum Of Array After K Negations.go

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+package leetcode
+
+import "sort"
+
+func largestSumAfterKNegations(A []int, K int) int {
+	sort.Ints(A)
+	minIdx := 0
+	for i := 0; i < K; i++ {
+		A[minIdx] = -A[minIdx]
+		if minIdx == len(A)-1 || A[minIdx+1] < A[minIdx] {
+			minIdx++
+		}
+	}
+	sum := 0
+	for _, a := range A {
+		sum += a
+	}
+	return sum
+}

Algorithms/1005. Maximize Sum Of Array After K Negations/1005. Maximize Sum Of Array After K Negations_test.go

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+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1005 struct {
+	para1005
+	ans1005
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1005 struct {
+	one []int
+	two int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1005 struct {
+	one int
+}
+
+func Test_Problem1005(t *testing.T) {
+
+	qs := []question1005{
+
+		question1005{
+			para1005{[]int{4, 2, 3}, 1},
+			ans1005{5},
+		},
+		question1005{
+			para1005{[]int{3, -1, 0, 2}, 3},
+			ans1005{6},
+		},
+
+		question1005{
+			para1005{[]int{2, -3, -1, 5, -4}, 2},
+			ans1005{13},
+		},
+
+		question1005{
+			para1005{[]int{-2, 9, 9, 8, 4}, 5},
+			ans1005{32},
+		},
+
+		question1005{
+			para1005{[]int{5, -7, -8, -3, 9, 5, -5, -7}, 7},
+			ans1005{49},
+		},
+
+		question1005{
+			para1005{[]int{5, 6, 9, -3, 3}, 2},
+			ans1005{20},
+		},
+
+		question1005{
+			para1005{[]int{8, -7, -3, -9, 1, 9, -6, -9, 3}, 8},
+			ans1005{53},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1005------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1005, q.para1005
+		fmt.Printf("【input】:%v       【output】:%v\n", p, largestSumAfterKNegations(p.one, p.two))
+	}
+	fmt.Printf("\n\n\n")
+}

Algorithms/1005. Maximize Sum Of Array After K Negations/README.md

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+# [1005. Maximize Sum Of Array After K Negations](https://leetcode.com/problems/maximize-sum-of-array-after-k-negations/)
+
+## 题目
+
+Given an array A of integers, we must modify the array in the following way: we choose an i and replace A[i] with -A[i], and we repeat this process K times in total.  (We may choose the same index i multiple times.)
+
+Return the largest possible sum of the array after modifying it in this way.
+
+
+Example 1:
+
+```c
+Input: A = [4,2,3], K = 1
+Output: 5
+Explanation: Choose indices (1,) and A becomes [4,-2,3].
+```
+
+Example 2:
+
+```c
+Input: A = [3,-1,0,2], K = 3
+Output: 6
+Explanation: Choose indices (1, 2, 2) and A becomes [3,1,0,2].
+```
+
+Example 3:
+
+```c
+Input: A = [2,-3,-1,5,-4], K = 2
+Output: 13
+Explanation: Choose indices (1, 4) and A becomes [2,3,-1,5,4].
+```
+
+Note:
+
+- 1 <= A.length <= 10000
+- 1 <= K <= 10000
+- -100 <= A[i] <= 100
+
+## 题目大意
+
+将数组中的元素变成它的相反数，这种操作执行 K 次之后，求出数组中所有元素的总和最大。
+
+这一题可以用最小堆来做，构建最小堆，每次将最小的元素变成它的相反数。然后最小堆调整，再将新的最小元素变成它的相反数。执行 K 次以后求数组中所有的值之和就是最大值。
+
+这道题也可以用排序来实现。排序一次，从最小值开始往后扫，依次将最小值变为相反数。这里需要注意一点，负数都改变成正数以后，接着不是再改变这些变成正数的负数，而是接着改变顺序的正数。因为这些正数是比较小的正数。负数越小，变成正数以后值越大。正数越小，变成负数以后对总和影响最小。具体实现见代码。