[leetcode] 861. Score After Flipping Matrix

Description

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. 1 <= A.length <= 20
2. 1 <= A[0].length <= 20
3. A[i][j] is 0 or 1.

分析

题目的意思是：每个flip操作把该行或者列的所有0变成1，所有1变成0。然后每行看做一个二进制数，把所有行加起来作为返回结果。问怎么翻转结果才最大？

• 1所在的位置，决定了它对最终的得分的大小。

代码

class Solution {
public:
    int matrixScore(vector<vector<int>>& A) {
        int m=A.size();
        int n=A[0].size();
        int res=0;
        for(int i=0;i<n;i++){
            int col=0;
            for(int j=0;j<m;j++){
                col+=A[j][i]^A[j][0];
            }
            res+=max(col,m-col)*(1<<(n-i-1));
        }
        return res;
    }
};

参考文献

861. Score After Flipping Matrix
leetcode讲解–861. Score After Flipping Matrix

标签：Matrix,861,After,Flipping,int,matrix,row
来源： https://blog.csdn.net/w5688414/article/details/94459041