https://leetcode.com/problems/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 to0

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.

題解：

class Solution {
public:
    int matrixScore(vector<vector<int>>& A) {
    	if (A.size() == 1){
    		return 1;
    	}
    	int ans = 0;
    	int row = A.size();
    	int col = A[0].size();
        //cout << A.size() << A[0].size();
    	for (int i = 0; i < row; i++){
    		if (A[i][0] == 0){
    			change_row(A, i);
    		}
    	}
		for (int i = 1; i < col; i++){
            int one = 0, zero = 0;
			for (int j = 0; j < row; j++){
				if (A[j][i] == 0){
					zero++;
				}
				else {
					one++;
				}
			}
            if (one < zero){
				change_column(A, i);
			}
		}
		for (int i = 0; i < row; i++){
			for (int j = 0; j < col; j++){
				ans += A[i][j] * pow(2, col - 1 - j);
			}
		}   
		return ans;
    }
    void change_row(vector<vector<int>>& A, int x){
    	for (int i = 0; i < A[x].size(); i++){
    		if (A[x][i] == 0){
    			A[x][i] = 1;
    		}
    		else{
    			A[x][i] = 0;
    		}
    	}
    }
    void change_column(vector<vector<int>>& A, int x){
    	for (int i = 0; i < A.size(); i++){
    		if (A[i][x] == 0){
    			A[i][x] = 1;
    		}
    		else{
    			A[i][x] = 0;
    		}
    	}
    }
};