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:
Input: A = [4,2,3], K = 1 Output: 5 Explanation: Choose indices (1,) and A becomes [4,-2,3].
Example 2:
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:
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 <= 100001 <= K <= 10000-100 <= A[i] <= 100
给定一个整数数组 A，我们只能用以下方法修改该数组：我们选择某个个索引
i 并将 A[i] 替换为 -A[i]，然后总共重复这个过程 K 次。（我们可以多次选择同一个索引 i。）
以这种方式修改数组后，返回数组可能的最大和。
示例 1：
输入：A = [4,2,3], K = 1 输出：5 解释：选择索引 (1,) ，然后 A 变为 [4,-2,3]。
示例 2：
输入：A = [3,-1,0,2], K = 3 输出：6 解释：选择索引 (1, 2, 2) ，然后 A 变为 [3,1,0,2]。
示例 3：
输入：A = [2,-3,-1,5,-4], K = 2 输出：13 解释：选择索引 (1, 4) ，然后 A 变为 [2,3,-1,5,4]。
提示：
1 <= A.length <= 100001 <= K <= 10000-100 <= A[i] <= 100
Runtime: 44 ms
Memory Usage: 19.1 MB
1 class Solution {
2 func largestSumAfterKNegations(_ A: [Int], _ K: Int) -> Int {
3 var A = A
4 var K = K
5 A.sort()
6 for i in 0..<A.count
7 {
8 if K > 0 && A[i] < 0
9 {
10 A[i] = -A[i]
11 K -= 1
12 }
13 }
14 A.sort()
15 if K % 2 == 1
16 {
17 A[0] = -A[0]
18 }
19 var ret:Int = 0
20 for a in A
21 {
22 ret += a
23 }
24 return ret
25 }
26 }
原文地址：https://www.cnblogs.com/strengthen/p/10504835.html
时间： 2024-10-08 16:34:35