LeetCode 861. Score After Flipping Matrix

最新推荐文章于 2020-12-08 20:08:04 发布

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本文介绍了一种解决LeetCode上“翻转矩阵后的最大得分”问题的方法。通过行翻转确保每一行的第一个元素为1，然后对于剩余的每一位，统计最优的列翻转策略以最大化整个矩阵的二进制数值总和。

题目链接：https://leetcode.com/problems/score-after-flipping-matrix/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.

题目解析：将各个数看作矩阵，将每个数按照数位来分，显然每个数的第一位是要保证都为1的（因为它比 A[i][1] + .. + A[i][m-1] 更大），分成两步——

先进行行翻转，让A[i][0] = 0，此时 A[i][j]==1 的条件为A[i][j] == A[i][0]；
后面的数位从列的角度来看，每列代表一个数位，要让每列有尽可能多的1，比较该列翻转与否中1的个数就可以获得较大值，相加即可。

代码如下：0ms Accepted beating 100%

class Solution {
public:
    int matrixScore(vector<vector<int>>& A) {
        int n = A.size(), m = A[0].size();
        int ans = (1 << (m - 1)) * n;
        for (int i = 1; i < m; i++)
        {
            int num = 0;
            for (int j = 0; j < n; j++)
                num += (A[j][i] == A[j][0]);
            ans += max(num, n - num) * (1 << (m - i - 1));
        }
        return ans;
    }
};