LeetCode 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).

The above rectangle (with the red border) is defined by (row1, col1) = (2, 1) and (row2, col2) = (4, 3), which contains sum = 8.

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:

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

Solution: can be simplified to add one column and one row to dp(dp[row+1][column+1] )

 1 public class NumMatrix {
 2     private int[,] dp {get;set;}
 3     public NumMatrix(int[,] matrix) {
 4         int r = matrix.GetLength(0);
 5         int c = matrix.GetLength(1);
 6         if(r>0&&c>0)
 7         {
 8             dp = new int[r, c];
 9             for(int i=0; i<r;i++)
10             {
11                 for(int j=0;j<c; j++)
12                 {
13                     if(i==0&&j==0)
14                     {
15                         dp[0,0]=matrix[0,0];
16                     }
17                     else if(i==0)
18                     {
19                         dp[0,j]=dp[0,j-1]+matrix[0,j];
20                     }
21                     else if(j==0)
22                     {
23                         dp[i,0] = dp[i-1,0]+matrix[i,0];
24                     }
25                     else
26                     {
27                         dp[i,j] = dp[i-1,j]+dp[i, j-1]-dp[i-1,j-1]+matrix[i,j];
28                     }
29                 }
30             }
31         }
32
33     }
34
35     public int SumRegion(int row1, int col1, int row2, int col2) {
36         if(row1==0 && col1==0)
37         {
38             return dp[row2,col2];
39         }
40         else if(col1==0)
41         {
42             return dp[row2, col2]-dp[row1-1, col2];
43         }
44         else if(row1==0)
45         {
46             return dp[row2, col2]-dp[row2, col1-1];
47         }
48         else
49         {
50             return dp[row2, col2]-dp[row1-1, col2]-dp[row2, col1-1]+dp[row1-1, col1-1];
51         }
52     }
53 }