861. Score After Flipping Matrix

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.

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

一个由0和1构成的矩阵，每次可以选择任意一行或列，行或列的数字0变成1,1变成0.

每一行代表一个二进制数，求能够成的这些二进制数和的最大值

 

由二进制转十进制原理，1的位置越靠前越好，所以第一步，把每一行的第一个数变成1.

第二步，对除了第一列外的每一列的1的个数做统计，如果翻转后能增加1的个数，就翻转该列。

 

class Solution {
public:
    int matrixScore(vector<vector<int>>& A) {
        int row = A.size();
        int col = A[0].size();
        vector<int> cnt(col, 0);
        for (int i = 0; i < row; ++i) {
            if (A[i][0] == 0) {
                for (int j = 0; j < col; ++j) {
                    A[i][j] = 1 - A[i][j];
                    if (A[i][j] == 1)   ++cnt[j];
                }
            }
            else {
                for (int j = 0; j < col; ++j) {
                    if (A[i][j] == 1)  ++cnt[j];
                }
            }
        }
        for (int j = 1; j < col; ++j) {
            if (cnt[j] * 2 < row) {
                for (int i = 0; i < row; ++i)
                    A[i][j] = 1 - A[i][j];
            }
        }
        int sum = 0;
        int v = 0;;
        int res = 0;
        for (int i = 0; i < row; ++i) {
            sum = 0;
            v = 1;
            for (int j = col - 1; j >= 0; --j) {
                sum += A[i][j] * v;
                v = v * 2;
            }
            res += sum;    
        }
        return res;   
    }
};