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提示

给你一个下标从 0 开始的 m x n 二进制矩阵 grid 。

我们按照如下过程,定义一个下标从 0 开始的 m x n 差值矩阵 diff :

  • 令第 i 行一的数目为 onesRowi 。
  • 令第 j 列一的数目为 onesColj 
  • 令第 i 行零的数目为 zerosRowi 。
  • 令第 j 列零的数目为 zerosColj 。
  • diff[i][j] = onesRowi + onesColj - zerosRowi - zerosColj

请你返回差值矩阵 diff 。

 

示例 1:

输入:grid = [[0,1,1],[1,0,1],[0,0,1]]
输出:[[0,0,4],[0,0,4],[-2,-2,2]]
解释:
- diff[0][0] = onesRow0 + onesCol0 - zerosRow0 - zerosCol0 = 2 + 1 - 1 - 2 = 0 
- diff[0][1] = onesRow0 + onesCol1 - zerosRow0 - zerosCol1 = 2 + 1 - 1 - 2 = 0 
- diff[0][2] = onesRow0 + onesCol2 - zerosRow0 - zerosCol2 = 2 + 3 - 1 - 0 = 4 
- diff[1][0] = onesRow1 + onesCol0 - zerosRow1 - zerosCol0 = 2 + 1 - 1 - 2 = 0 
- diff[1][1] = onesRow1 + onesCol1 - zerosRow1 - zerosCol1 = 2 + 1 - 1 - 2 = 0 
- diff[1][2] = onesRow1 + onesCol2 - zerosRow1 - zerosCol2 = 2 + 3 - 1 - 0 = 4 
- diff[2][0] = onesRow2 + onesCol0 - zerosRow2 - zerosCol0 = 1 + 1 - 2 - 2 = -2
- diff[2][1] = onesRow2 + onesCol1 - zerosRow2 - zerosCol1 = 1 + 1 - 2 - 2 = -2
- diff[2][2] = onesRow2 + onesCol2 - zerosRow2 - zerosCol2 = 1 + 3 - 2 - 0 = 2

示例 2:

输入:grid = [[1,1,1],[1,1,1]]
输出:[[5,5,5],[5,5,5]]
解释:
- diff[0][0] = onesRow0 + onesCol0 - zerosRow0 - zerosCol0 = 3 + 2 - 0 - 0 = 5
- diff[0][1] = onesRow0 + onesCol1 - zerosRow0 - zerosCol1 = 3 + 2 - 0 - 0 = 5
- diff[0][2] = onesRow0 + onesCol2 - zerosRow0 - zerosCol2 = 3 + 2 - 0 - 0 = 5
- diff[1][0] = onesRow1 + onesCol0 - zerosRow1 - zerosCol0 = 3 + 2 - 0 - 0 = 5
- diff[1][1] = onesRow1 + onesCol1 - zerosRow1 - zerosCol1 = 3 + 2 - 0 - 0 = 5
- diff[1][2] = onesRow1 + onesCol2 - zerosRow1 - zerosCol2 = 3 + 2 - 0 - 0 = 5

 

提示:

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 105
  • 1 <= m * n <= 105
  • grid[i][j] 要么是 0 ,要么是 1
通过次数
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提示 1
You need to reuse information about a row or a column many times. Try storing it to avoid computing it multiple times.

提示 2
Use an array to store the number of 1’s in each row and another array to store the number of 1’s in each column. Once you know the number of 1’s in each row or column, you can also easily calculate the number of 0’s.


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行 1,列 1
grid =
[[0,1,1],[1,0,1],[0,0,1]]
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