【LeetCode 每日一题】2482. 行和列中一和零的差值

2482. 行和列中一和零的差值

难度：中等

时间：2023/06/18

---

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 <= 10^5
• 1 <= m * n <= 10^5
• grid[i][j] 要么是 0 ，要么是 1 。

---

解题思路：

按照题目大意模拟即可

class Solution {
    public int[][] onesMinusZeros(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        int[] oneRow = new int[m];
        int[] oneCol = new int[n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int cur = grid[i][j];
                if (cur == 1) {
                    oneRow[i]++;
                    oneCol[j]++;
                }
            }
        }
        int[][] ans = new int[m][n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                ans[i][j] = oneRow[i] + oneCol[j] - (m - oneRow[i]) - (n - oneCol[j]);
            }
        }
        return ans;
    }
​
}