README_CN.md

2482. 行和列中一和零的差值

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 。

题解 (Rust)

1. 题解

impl Solution {
    pub fn ones_minus_zeros(grid: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
        let m = grid.len();
        let n = grid[0].len();
        let mut ones_row = vec![0; m];
        let mut ones_col = vec![0; n];
        let mut diff = vec![vec![0; n]; m];

        for i in 0..m {
            for j in 0..n {
                ones_row[i] += grid[i][j];
                ones_col[j] += grid[i][j];
            }
        }

        for i in 0..m {
            for j in 0..n {
                diff[i][j] = 2 * ones_row[i] + 2 * ones_col[j] - (m + n) as i32;
            }
        }

        diff
    }
}