1005 Maximize Sum Of Array After K Negations K 次取反后最大化的數(shù)組和
Description:
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:
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 <= 10000
1 <= K <= 10000
-100 <= A[i] <= 100
題目描述:
給定一個(gè)整數(shù)數(shù)組 A，我們只能用以下方法修改該數(shù)組：我們選擇某個(gè)個(gè)索引 i 并將 A[i] 替換為 -A[i]，然后總共重復(fù)這個(gè)過(guò)程 K 次。（我們可以多次選擇同一個(gè)索引 i。）
以這種方式修改數(shù)組后，返回?cái)?shù)組可能的最大和。
示例 :
示例 1：
輸入：A = [4,2,3], K = 1
輸出：5
解釋：選擇索引 (1,) ，然后 A 變?yōu)?[4,-2,3]。
示例 2：
輸入：A = [3,-1,0,2], K = 3
輸出：6
解釋：選擇索引 (1, 2, 2) ，然后 A 變?yōu)?[3,1,0,2]。
示例 3：
輸入：A = [2,-3,-1,5,-4], K = 2
輸出：13
解釋：選擇索引 (1, 4) ，然后 A 變?yōu)?[2,3,-1,5,4]。
提示：
1 <= A.length <= 10000
1 <= K <= 10000
-100 <= A[i] <= 100
思路:
- 每次取最小的負(fù)數(shù)(絕對(duì)值最大的負(fù)數(shù))反轉(zhuǎn), 如果數(shù)組中全是正數(shù), 取最小的正數(shù)反轉(zhuǎn)即可
由于數(shù)組中的元素的絕對(duì)值均小于等于100, 可以將數(shù)組中的元素映射到一個(gè)長(zhǎng)度為 201的數(shù)組中
時(shí)間復(fù)雜度O(kn), 空間復(fù)雜度O(1) - 每次先排序, 然后將最小的數(shù)取反
時(shí)間復(fù)雜度O(knlgn), 空間復(fù)雜度O(1)
代碼:
C++:
class Solution
{
public:
int largestSumAfterKNegations(vector<int>& A, int K)
{
int num[201]{0};
for (auto a : A) ++num[a + 100];
int i = 0, result = 0;
while (K--)
{
while (!num[i]) ++i;
--num[i];
++num[200 - i];
if (i > 100) i = 200 - i;
}
for (; i < 201; i++) result += (i - 100) * num[i];
return result;
}
};
Java:
class Solution {
public int largestSumAfterKNegations(int[] A, int K) {
int num[] = new int[201];
for (int a : A) num[a + 100]++;
int i = 0, result = 0;
while (K-- > 0) {
while (num[i] == 0) i++;
num[i]--;
num[200 - i]++;
if (i > 100) i = 200 - i;
}
for (; i < 201; i++) result += (i - 100) * num[i];
return result;
}
}
Python:
class Solution:
def largestSumAfterKNegations(self, A: List[int], K: int) -> int:
for _ in range(K):
A.sort()
A[0] = -A[0]
return sum(A)
