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 0
s to 1
s, and all 1
s to 0
s.
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]
is0
or1
.
题目理解:
给定一个数组,只含有0和1.可以对这个数组的某行或者某列进行跳变,即0变1,1变0,最后将每一行组成一个二进制数,返回这些数的可能的最大的和
解题思路:
既然是二进制数,那左边的位代表的数值就大,所以我们要尽量将数组左边的数字变为1.又因为可以对某一行进行跳变,因此我们总是可以将第一列的全部数字变为1,而且我们必须这样做,因为最左边的位有最大的数值。在将第一列变为1之后我们就不再对行做跳变了,因为我们需要保持第一列全为1.我们之后只对每一列进行跳变,使得每一列中都有尽量多的1,最终将数字相加就好了。相加的时候其实不需要算出来每一行代表的十进制数,我们加上每一位代表的数就行了
代码如下:
class Solution {
public int matrixScore(int[][] A) {
int row = A.length;
if(row == 0)
return 0;
int col = A[0].length;
int res = row * (1 << (col - 1));
for(int i = 0; i < row; i++) {
if(A[i][0] == 1)
continue;
for(int j = 0; j < col; j++) {
A[i][j] = (A[i][j] + 1) % 2;
}
}
for(int j = 1; j < col; j++) {
int ct = 0;
for(int i = 0; i < row; i++) {
ct += A[i][j];
}
ct = Math.max(ct, row - ct);
res += ct * (1 << (col - 1 - j));
}
return res;
}
}