Score After Flipping Matrix

最新推荐文章于 2025-06-05 01:15:00 发布

Charles-Liu

最新推荐文章于 2025-06-05 01:15:00 发布

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分类专栏： leetcode 贪心

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本文介绍了解决LeetCode上ScoreAfterFlippingMatrix问题的方法。该问题要求通过翻转矩阵的行或列来最大化矩阵的分数，其中矩阵由0和1组成，每行被视为一个二进制数。解决方案首先确保每行的第一个元素为1，然后优化每列以增加总的二进制值。

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[leetcode]Score After Flipping Matrix

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

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.

Solution

// 想法是先通过toggle每一个row让每一行的第一个数字是1，然后逐行toggle后面的每一列，让每一列的1最多
class Solution {
public:
  int matrixScore(vector<vector<int>>& A) {
    int row = A.size();
    int col = A[0].size();
    int cnt = 0;
    for (int i = 0; i < row; i++) {
      if (A[i][0] == 0) {
        for (int j = 0; j < col; j++) {
          A[i][j] = !A[i][j];
        }
      }
      cnt += (1 << (col-1));
    }
    for (int j = 1; j < col; j++) {
      int zeros = 0, ones = 0;
      for (int i = 0; i < row; i++) {
        if (A[i][j] == 0) zeros++;
        else ones++;
      }
      int max_val = max(zeros, ones);
      cnt += (1 << (col-j-1))*max_val;
    }
    return cnt;
  }
};

思路：看代码注释~