[LeetCode] 2536. Increment Submatrices by One

You are given a positive integer n, indicating that we initially have an n x n 0-indexed integer matrix mat filled with zeroes.

You are also given a 2D integer array query. For each query[i] = [row1i, col1i, row2i, col2i], you should do the following operation:

• Add 1 to every element in the submatrix with the top left corner (row1i, col1i) and the bottom right corner (row2i, col2i). That is, add 1 to mat[x][y] for for all row1i <= x <= row2i and col1i <= y <= col2i.

Return the matrix mat after performing every query.

Example 1:

Input: n = 3, queries = [[1,1,2,2],[0,0,1,1]]
Output: [[1,1,0],[1,2,1],[0,1,1]]
Explanation: The diagram above shows the initial matrix, the matrix after the first query, and the matrix after the second query.
- In the first query, we add 1 to every element in the submatrix with the top left corner (1, 1) and bottom right corner (2, 2).
- In the second query, we add 1 to every element in the submatrix with the top left corner (0, 0) and bottom right corner (1, 1).

Example 2:

Input: n = 2, queries = [[0,0,1,1]]
Output: [[1,1],[1,1]]
Explanation: The diagram above shows the initial matrix and the matrix after the first query.
- In the first query we add 1 to every element in the matrix.

Constraints:

• 1 <= n <= 500
• 1 <= queries.length <= 104
• 0 <= row1i <= row2i < n
• 0 <= col1i <= col2i < n

子矩阵元素加 1。

给你一个正整数 n ，表示最初有一个 n x n 、下标从 0 开始的整数矩阵 mat ，矩阵中填满了 0 。

另给你一个二维整数数组 query 。针对每个查询 query[i] = [row1i, col1i, row2i, col2i] ，请你执行下述操作：

找出 左上角 为 (row1i, col1i) 且 右下角 为 (row2i, col2i) 的子矩阵，将子矩阵中的 每个元素 加 1 。也就是给所有满足 row1i <= x <= row2i 和 col1i <= y <= col2i 的 mat[x][y] 加 1 。
返回执行完所有操作后得到的矩阵 mat 。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/increment-submatrices-by-one
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思路是二维差分。二维差分可以解决类似这样的问题，比如下图（引用）

在一个二维矩阵里，

(a) 如果你想把桔黄色的部分都 + 1，那么就只需要在左上角这个点 + 1，然后遍历整个矩阵，令 grid[i][j] += grid[i][j - 1] 即可

如果你只想让 (b) 图中蓝色的部分都 + 1，那么你需要在图 (c) 的这些位置做加或减的操作，然后再求整个矩阵的前缀和。这道题也是一样的思路，只是注意，需要修改的部分的右下角的坐标不能越界。

时间O(n^2)

空间O(n^2)

Java实现

class Solution {
    public int[][] rangeAddQueries(int n, int[][] queries) {
        int[][] grid = new int[n][n];
        for (int[] q : queries) {
            int r1 = q[0];
            int c1 = q[1];
            int r2 = q[2];
            int c2 = q[3];
            for (int i = r1; i <= r2; i++) {
                grid[i][c1]++;
                // 注意不能越界
                if (c2 + 1 < n) {
                    grid[i][c2 + 1]--;
                }
            }
        }

        for (int i = 0; i < n; i++) {
            for (int j = 1; j < n; j++) {
                grid[i][j] += grid[i][j - 1];
            }
        }
        return grid;
    }
}

 

LeetCode 题目总结