Leetcode 861. Score after flipping a matrix

There is a two-dimensional matrix A where each element has a value of 0 or 1 .

Move refers to selecting any row or column and converting each value in that row or column: Change all 0 1 to, change all to 1 0 .

After making any number of moves, each row of the matrix is interpreted as a binary number, and the Matrix's score is the sum of those numbers.

Returns the highest possible score.

Example:

Input: [[[0,0,1,1],[1,0,1,0],[1,1,0,0]] Output: 39 Explanation: Convert to [[1,1,1,1],[1,0,0,1],[1,1,1,1]]0b1111 + 0b1001 + 0b1111 = 15 + 9 + 15 = 39

A simple greedy strategy. Make sure that each row starts with 0, and then make sure that each column is as much as 1.

classSolution { Public:    intMatrixscore (vector<vector<int>>&A) {if(a.size () = =0|| a[0].size () = =0) {            return 0; }        introw = A.size (), column = a[0].size (), sum =0;  for(inti =0; i < row; i++) {            if(a[i][0] ==0) {                 for(intj =0; J < column; J + +) {A[i][j]=1-A[i][j]; }            }        }         for(intj =0; J < column; J + +) {            intZero =0, one =0;  for(inti =0; i < row; i++) {                if(A[i][j] = =0) {Zero++; } Else{ One++; }            }            if(Zero >One ) {                 for(inti =0; i < row; i++) {A[i][j]=1-A[i][j]; }            }        }         for(inti =0; i < row; i++) {             for(intj =0; J < column; J + +) {sum+ = a[i][j] * POW (2, Column-j-1); }        }        returnsum; }};

Leetcode 861. Score after flipping a matrix