Leetcode Range Sum Query 2D - Immutable

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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
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:

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.

解题思路：用每sp[i][j]表示从sp[i][0]到sp[i][j]累加和，
求解时，只需要将矩阵坐在行的sp[i+1][col2+1]-sp[i][col1]值相加即可；
public class RangeSumQuery2DImmutable
{
int[][] sp;

public void NumMatrix(int[][] matrix)
{
    sp=new int[matrix.length+1][matrix[0].length+1];
    sp[0][0]=0;
    for(int i=0;i<matrix.length;i++)
    {
        sp[i][0]=0;
    }
    for(int j=0;j<matrix[0].length;j++)
    {
        sp[0][j]=0;
    }

    for(int i=1;i<=matrix.length;i++)
    {
        for(int j=1;j<=matrix[0].length;j++)
        {
            sp[i][j]=sp[i][j-1]+matrix[i-1][j-1];
        }
    }

}

public int sumRegion(int row1, int col1, int row2, int col2)
{
    int result=0;
    for(int i=row1;i<=row2;i++)
    {
        result=result+sp[i+1][col2+1]-sp[i][col1];
    }
    return result;
}

}