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2482. 行和列中一和零的差值

English Version

题目描述

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 。

解法

方法一：模拟

根据题意模拟即可。

时间复杂度 $O(m \times n)$，忽略答案的空间消耗，空间复杂度 $O(m + n)$。其中 $m$ 和 $n$ 分别为矩阵的行数和列数。

Python3

class Solution:
    def onesMinusZeros(self, grid: List[List[int]]) -> List[List[int]]:
        m, n = len(grid), len(grid[0])
        rows = [0] * m
        cols = [0] * n
        for i, row in enumerate(grid):
            for j, v in enumerate(row):
                rows[i] += v
                cols[j] += v
        diff = [[0] * n for _ in range(m)]
        for i, r in enumerate(rows):
            for j, c in enumerate(cols):
                diff[i][j] = r + c - (n - r) - (m - c)
        return diff

Java

class Solution {
    public int[][] onesMinusZeros(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        int[] rows = new int[m];
        int[] cols = new int[n];
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int v = grid[i][j];
                rows[i] += v;
                cols[j] += v;
            }
        }
        int[][] diff = new int[m][n];
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                diff[i][j] = rows[i] + cols[j] - (n - rows[i]) - (m - cols[j]);
            }
        }
        return diff;
    }
}

C++

class Solution {
public:
    vector<vector<int>> onesMinusZeros(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        vector<int> rows(m);
        vector<int> cols(n);
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int v = grid[i][j];
                rows[i] += v;
                cols[j] += v;
            }
        }
        vector<vector<int>> diff(m, vector<int>(n));
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                diff[i][j] = rows[i] + cols[j] - (n - rows[i]) - (m - cols[j]);
            }
        }
        return diff;
    }
};

Go

func onesMinusZeros(grid [][]int) [][]int {
	m, n := len(grid), len(grid[0])
	rows := make([]int, m)
	cols := make([]int, n)
	diff := make([][]int, m)
	for i, row := range grid {
		diff[i] = make([]int, n)
		for j, v := range row {
			rows[i] += v
			cols[j] += v
		}
	}
	for i, r := range rows {
		for j, c := range cols {
			diff[i][j] = r + c - (n - r) - (m - c)
		}
	}
	return diff
}

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