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861. Score After Flipping Matrix

时间:2019-01-22 20:41:09      阅读:157      评论:0      收藏:0      [点我收藏+]

标签:max   erp   ogg   not   input   bsp   pos   nsis   ssi   

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. 1 <= A.length <= 20
  2. 1 <= A[0].length <= 20
  3. A[i][j] is 0 or 1.

 

Approach #1: C++.

class Solution {
public:
    int matrixScore(vector<vector<int>>& A) {
        int r = A.size(), c = A[0].size();
        int ans = 0;
        for (int i = 0; i < c; ++i) {
            int col = 0;
            for (int j = 0; j < r; ++j)
                col += A[j][i] ^ A[j][0];       //相同为0, 不同为1。
            ans += max(col, r - col) * (1 << (c - 1 - i));
        }
        return ans;
    }
};

  

 

861. Score After Flipping Matrix

标签:max   erp   ogg   not   input   bsp   pos   nsis   ssi   

原文地址:https://www.cnblogs.com/ruruozhenhao/p/10305862.html

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