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# 題目敘述
You are given an m x n binary matrix grid .
A move consists of choosing any row or column and toggling each value in that row or column (i.e., changing all 0 's to 1 's, and all 1 's to 0 's).
Every row of the matrix is interpreted as a binary number, and the score of the matrix is the sum of these numbers.
Return the highest possible score after making any number of moves (including zero moves).
# Example 1
Input: grid = [[0,0,1,1],[1,0,1,0],[1,1,0,0]]
Output: 39
Explanation: 0b1111 + 0b1001 + 0b1111 = 15 + 9 + 15 = 39
# Example 2
Input: grid = [[0]]
Output: 1
# 解題思路
# Solution
class Solution { | |
public int matrixScore(int[][] grid) { | |
int row = grid.length, col = grid[0].length, ans = 0; | |
for(int i = 0; i < row; i++){ | |
if(grid[i][0] == 0) Flipping(grid, col, i, true); | |
} | |
for(int i = 0; i < col; i++){ | |
int zero = 0; | |
for(int j = 0; j < row; j++) if(grid[j][i] == 0) zero++; | |
if(row / 2 < zero) Flipping(grid, row, i, false); | |
} | |
for(int i = 0; i < row; i++){ | |
int tmp = 0; | |
for(int j = 0; j < col; j++) tmp = tmp * 2 + grid[i][j]; | |
ans += tmp; | |
} | |
return ans; | |
} | |
private void Flipping(int[][] grid, int len, int i, boolean isRow){ | |
for(int j = 0; j < len; j++){ | |
if(isRow) grid[i][j] = grid[i][j] ^ 1; | |
else grid[j][i] = grid[j][i] ^ 1; | |
} | |
} | |
} |
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