861. Score After Flipping Matrix

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

解题思路：

对于一个矩阵有两种变换，行和列上面的取反。

对于行变换，若首位数字不为1时，将进行行变化，此时整体数值变大。

对于列变换，从第二列开始，统计每列的1和0的个数，若0的个数多余1的个数时，进行列变换，此时整体数值变大。

代码：

class Solution {
public:
    int matrixScore(vector<vector<int>>& A) {
        for (auto & vec : A) {
            if (!vec[0])
                togging(vec);
        }
        for (int i = 1; i < A[0].size() ; ++i){
            int zero = 0;
            int one = 0;
            for (int j = 0; j < A.size(); ++j){
                if (A[j][i])
                    one++;
                else
                    zero++;
            }
            if (zero>one){
                for (int j = 0; j < A.size(); j++){
                    A[j][i] = !A[j][i];
                }
            }
        }
        int num = 0;
        for (auto vec :A) {
            for (int i = 0; i < vec.size(); i++){
                num += vec[i] *pow(2, (vec.size()-1-i));
            }
        }
        return num;
    }
    void togging(vector<int>& vec){
        for(int i = 0; i <vec.size(); i++){
            vec[i] = !vec[i];
        }
    }
};