LeetCode #861. Score After Flipping Matrix

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本文探讨了如何通过切换矩阵的行和列来最大化矩阵的分数，解释了一种高效的算法实现，该算法能够确保每行的最高位为1，并通过统计各列中1的数量来进一步优化矩阵的总分数。

题目描述：

We have a two dimensional matrix A where each value is 0 or 1.

A move consists of choosing any row or column, and toggling each value in that row or column: changing all 0s to 1s, and all 1s to 0s.

After making any number of moves, every row of this matrix is interpreted as a binary number, and the score of the matrix is the sum of these numbers.

Return the highest possible score.

Example 1:

Input: [[0,0,1,1],[1,0,1,0],[1,1,0,0]]
Output: 39
Explanation:
Toggled to [[1,1,1,1],[1,0,0,1],[1,1,1,1]].
0b1111 + 0b1001 + 0b1111 = 15 + 9 + 15 = 39

Note:

1 <= A.length <= 20
1 <= A[0].length <= 20
A[i][j] is 0 or 1.

class Solution {
public:
    int matrixScore(vector<vector<int>>& A) {
        int m=A.size(), n=A[0].size();
        // 每一行的最高位一定要设置成1
        int result=m*(1<<(n-1));
        for(int j=1;j<n;j++)
        {
            int count=0;
            // 翻转完每一行之后，再翻转每一列，将当前元素和最高位比较
            // 如果相同的话，说明这一行翻转后当前元素也是1，否则是0
            // 统计当前列一共有多少个1和0，那么选个数较大的一组设定为1
            for(int i=0;i<m;i++)
                if(A[i][j]==A[i][0]) count++;
            result+=max(count,m-count)*(1<<(n-j-1));
        }
        return result;
    }
};