304. Range Sum Query 2D - Immutable

Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).

Range Sum Query 2D

Example:
Given matrix = [
[3, 0, 1, 4, 2],
[5, 6, 3, 2, 1],
[1, 2, 0, 1, 5],
[4, 1, 0, 1, 7],
[1, 0, 3, 0, 5]
]
sumRegion(2, 1, 4, 3) -> 8
sumRegion(1, 1, 2, 2) -> 11
sumRegion(1, 2, 2, 4) -> 12

Note:
You may assume that the matrix does not change.
There are many calls to sumRegion function.
You may assume that row1 ≤ row2 and col1 ≤ col2.

public class NumMatrix {
 
    int sum[][] = null;
    
    public NumMatrix(int[][] matrix) {
        if(matrix.length == 0){
            return;
        }
        sum = new int[matrix.length][matrix[0].length];
        sum[0][0] = matrix[0][0];
        
        for(int i = 1; i< matrix.length; i++){
            sum[i][0] = matrix[i][0] + sum[i-1][0];
        }
        
        for(int j = 1; j< matrix[0].length; j++){
            sum[0][j] = matrix[0][j] + sum[0][j-1];
        }
        
        for(int i = 1 ; i < matrix.length; i++){
            for(int j = 1; j < matrix[0].length; j++){
                sum[i][j] = sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1] + matrix[i][j];
                            
            }
        }
    }

    public int sumRegion(int row1, int col1, int row2, int col2) {
        if(sum == null){
            return 0;
        }
        if (row1 == 0 && col1 == 0) {
            return sum[row2][col2];
        } else if (row1 == 0) {
            return sum[row2][col2] - sum[row2][col1 - 1];
        } else if (col1 == 0) {
            return sum[row2][col2] - sum[row1 - 1][col2];
        } else {
            return sum[row2][col2] - sum[row1 - 1][col2] - 
                (sum[row2][col1 - 1] - sum[row1 - 1][col1 - 1]);
        }
    }
}