LeetCode 1975 Maximum Matrix Sum (贪心)

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该博客讨论了一种优化策略，通过选择矩阵中相邻元素并乘以-1来最大化矩阵元素之和。问题的核心在于确定何时通过操作将负数转换为正数以增加总和。文章分析了当矩阵中没有0且负数个数为奇数时的特殊情况，并提供了高效的解决方案。代码示例展示了如何实现这一算法，以4ms的时间复杂度达到高效率。

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
-10^5 <= matrix[i][j] <= 10^5

题目链接：https://leetcode.com/problems/maximum-matrix-sum/

题目大意：一次操作可任选两相邻格子并让其值都乘-1，可操作无数次，求矩阵和的最大值

题目分析：因为可以操作无数次，若矩阵中有0或负数的个数是偶数，最终必能让所有值非负（0可以用来将任何负数转为正数），故只有矩阵中没有0且有奇数个负数时需要额外考虑，此时也只需看负数的最大值和正数的最小值的绝对值谁更小，减去即可

4ms，时间击败94.71%

class Solution {
    public long maxMatrixSum(int[][] matrix) {
        int n = matrix.length, negCnt = 0;
        int negMa = Integer.MIN_VALUE, posMi = Integer.MAX_VALUE;
        long ans = 0;
        boolean hasZero = false;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] < 0) {
                    negCnt++;
                    if (negMa < matrix[i][j]) {
                        negMa = matrix[i][j];
                    }
                } else if (matrix[i][j] > 0) {
                    if (posMi > matrix[i][j]) {
                        posMi = matrix[i][j];
                    }
                } else {
                    hasZero = true;
                }
                ans += Math.abs(matrix[i][j]);
            }
        }
        if (!hasZero && negCnt % 2 == 1) {
            int extra = -negMa < posMi ? -negMa * 2 : posMi * 2;
            ans -= extra;
        }
        return ans;
    }
}