LeetCode每日一题(1975. Maximum Matrix Sum)

You are given an n x n integer matrix. You can do the following operation any number of times:

Choose any two adjacent elements of matrix and multiply each of them by -1.
Two elements are considered adjacent if and only if they share a border.

Your goal is to maximize the summation of the matrix’s elements. Return the maximum sum of the matrix’s elements using the operation mentioned above.

Example 1:

Input: matrix = [[1,-1],[-1,1]]
Output: 4

Explanation: We can follow the following steps to reach sum equals 4:

• Multiply the 2 elements in the first row by -1.
• Multiply the 2 elements in the first column by -1.

Example 2:

Input: matrix = [[1,2,3],[-1,-2,-3],[1,2,3]]
Output: 16

Explanation: We can follow the following step to reach sum equals 16:

• Multiply the 2 last elements in the second row by -1.

Constraints:

• n == matrix.length == matrix[i].length
• 2 <= n <= 250
• -105 <= matrix[i][j] <= 105

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这个题换个说法，其实就是， 我们可以任意的移动负号(-)， 当 2 个负号凑到一起可以消除。

如果有偶数个负数， 我们可以把它们都变成正数

如果有奇数个负数， 最终会剩余一个负数，但是因为我们可以自由移动负号， 我们可以把负号移动到绝对值最小的数上以获取最大的加和

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impl Solution {
    pub fn max_matrix_sum(matrix: Vec<Vec<i32>>) -> i64 {
        let mut negative_count = 0;
        let mut positives = Vec::new();
        for l in matrix {
            for v in l {
                if v < 0 {
                    negative_count += 1;
                }
                positives.push(v.abs() as i64);
            }
        }
        positives.sort();
        if negative_count % 2 == 1 {
            positives[0] = -positives[0];
        }
        positives.into_iter().sum()
    }
}