題目描述：

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.

解題思路：

對於一個矩陣有兩種變換，行和列上面的取反。

對於行變換，若首位數字不為1時，將進行行變化，此時整體數值變大。

對於列變換，從第二列開始，統計每列的1和0的個數，若0的個數多余1的個數時，進行列變換，此時整體數值變大。

代碼：

 1 class Solution {
 2 public:
 3     int matrixScore(vector<vector<int>>& A) {
 4         for (auto & vec : A) {
 5             if (!vec[0])
 6                 togging(vec);
 7         }
 8         for (int i = 1; i < A[0
 
].size() ; ++i){
 9             int zero = 0;
10             int one = 0;
11             for (int j = 0; j < A.size(); ++j){
12                 if (A[j][i])
13                     one++;
14                 else
15                     zero++;
16             }
17             if (zero>one){
18                 for (int j = 0; j < A.size(); j++){
19                     A[j][i] = !A[j][i];
20                 }
21             }
22         }
23         int num = 0;
24         for (auto vec :A) {
25             for (int i = 0; i < vec.size(); i++){
26                 num += vec[i] *pow(2, (vec.size()-1-i));
27             }
28         }
29         return num;
30     }
31     void togging(vector<int>& vec){
32         for(int i = 0; i <vec.size(); i++){
33             vec[i] = !vec[i];
34         }
35     }
36 };

861. Score After Flipping Matrix