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 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.
分析
题目的意思是:每个flip操作把该行或者列的所有0变成1,所有1变成0。然后每行看做一个二进制数,把所有行加起来作为返回结果。问怎么翻转结果才最大?
- 1所在的位置,决定了它对最终的得分的大小。
代码
class Solution {
public:
int matrixScore(vector<vector<int>>& A) {
int m=A.size();
int n=A[0].size();
int res=0;
for(int i=0;i<n;i++){
int col=0;
for(int j=0;j<m;j++){
col+=A[j][i]^A[j][0];
}
res+=max(col,m-col)*(1<<(n-i-1));
}
return res;
}
};
参考文献
861. Score After Flipping Matrix
leetcode讲解–861. Score After Flipping Matrix