861. Score After Flipping Matrix

#### QUESTION:

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

#### EXPLANATION:

2.从第二列开始，计算每一列0和1的数量，来确定是否要翻转这一列
3.将得到的结果进行累加

#### SOLUTION:

class Solution {
public int matrixScore(int[][] A) {
for(int i = 0;i<A.length;i++)
if(A[i][0]==0) matrixScoreRow(A,i);
for( int i =1;i<A[0].length;i++){
int count0 = 0;
for(int j = 0;j<A.length;j++) if(A[j][i]==0) count0++;
if(A.length-count0<count0) matrixScoreColum(A,i);
}
int result = 0;
for(int i = 0;i<A.length;i++)
result += matrixScoreRowBinary(A[i]);
return result;
}

public static int matrixScoreRowBinary(int[] A){
StringBuilder sb = new StringBuilder();
for(int i = 0;i<A.length;i++) sb.append(A[i]);
return Integer.parseInt(sb.toString(),2);
}

public static void matrixScoreRow(int[][] A,int row){
for(int i = 0;i<A[0].length;i++) A[row][i] = Math.abs(A[row][i]-1);
}

public static void matrixScoreColum(int[][] A,int colum){
for(int i = 0;i<A.length;i++) A[i][colum] = Math.abs(A[i][colum]-1);
}

}