LeetCode 1314. Matrix Block Sum

原题链接在这里：https://leetcode.com/problems/matrix-block-sum/

题目：

Given a m x n matrix mat and an integer k, return a matrix answer where each answer[i][j] is the sum of all elements mat[r][c] for:

• i - k <= r <= i + k,
• j - k <= c <= j + k, and
• (r, c) is a valid position in the matrix.

Example 1:

Input: mat = [[1,2,3],[4,5,6],[7,8,9]], k = 1
Output: [[12,21,16],[27,45,33],[24,39,28]]

Example 2:

Input: mat = [[1,2,3],[4,5,6],[7,8,9]], k = 2
Output: [[45,45,45],[45,45,45],[45,45,45]]

Constraints:

• m == mat.length
• n == mat[i].length
• 1 <= m, n, k <= 100
• 1 <= mat[i][j] <= 100

题解：

First have a range sum array nums.

nums[i][j] = mat[i - 1][j - 1] + nums[i][j - 1] + nums[i - 1][j] - nums[i - 1][j - 1]. 

For the block. we would to get the r1, c1, r2, c2.

Then calculate the block value mat[i][j] = nums[r2][c2] - nums[r1][c2] - nums[r2][c1] + nums[r1][c1].

Time Compleixity: O(m * n). m = mat.length. n = mat[0].length.

Space: O(m * n)

AC Java:

class Solution {
    public int[][] matrixBlockSum(int[][] mat, int k) {
        if(mat == null || mat.length == 0 || mat[0].length == 0){
            return mat;
        }
        
        int m = mat.length;
        int n = mat[0].length;
        int [][] nums = new int[m + 1][n + 1];
        for(int i = 1; i <= m; i++){
            for(int j = 1; j <= n; j++){
                nums[i][j] = mat[i - 1][j - 1] + nums[i - 1][j] + nums[i][j - 1] - nums[i - 1][j - 1];
            }
        }
        
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                int r1 = Math.max(0, i - k);
                int r2 = Math.min(m, i + k + 1);
                int c1 = Math.max(0, j - k);
                int c2 = Math.min(n, j + k + 1);
                mat[i][j] = nums[r2][c2] - nums[r1][c2] - nums[r2][c1] + nums[r1][c1];
            }
        }
        
        return mat;
    }
}

类似Range Sum Query 2D - Immutable.