LeetCode Range Sum Query 2D - Immutable

题目：

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:

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.

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题意：
就是给定四个值，分别代表一块区间内的左上的节点和右下的节点，然后计算这块区域中的所有节点的和。题目中给的结构是：

public class NumMatrix
{
public NumMatrix(int[][] matrix)
{

}
public int sumRegion(int row1,int col1,int row2,int col2)
{

}

从中我们可以发现，其实题目的意思是要我们能够采用类的定义的方法，用
// Your NumMatrix object will be instantiated and called as such:

// NumMatrix numMatrix = new NumMatrix(matrix);

// numMatrix.sumRegion(0, 1, 2, 3);

// numMatrix.sumRegion(1, 2, 3, 4);

来做，所以我们可以考虑新生成一个数组，matrixSums[][]，这个数组用来保存某一个节点到源节点的那一块区域的和。然后在下面的那个函数里，求这一块区域的和。

public class NumMatrix
{

public int[][] matrixSums;
public NumMatrix(int[][] matrix)
{
matrixSums = null;
if(matrix.length > 0 && matrix[0].length > 0)
{
matrixSums = new int[matrix.length][matrix[0].length];
for(int i = 0; i < matrix.length; i++)
{
for(int j = 0; j < matrix[0].length; j++)
{
if(i == 0)
{
if(j == 0)
matrixSums[i][j] = matrix[i][j];
else
matrixSums[i][j] = matrixSums[i][j-1] + matrix[i][j];
}
else
{
if(j == 0)
matrixSums[i][j] = matrixSums[i-1][j] + matrix[i][j];
else
matrixSums[i][j] = matrixSums[i-1][j] + matrixSums[i][j-1] + matrix[i][j] - matrixSums[i-1][j-1];
}
}
}
}
}
public int sumRegion(int row1,int col1,int row2,int col2)
{
if(matrixSums == null || row1 > row2 || col1 > col2)
return 0;
else if(row1 == 0 && col1 == 0)
return matrixSums[row2][col2];
else if(row1 == 0)
return matrixSums[row2][col2] - matrixSums[row2][col1-1];
else if(col1 == 0)
return matrixSums[row2][col2] - matrixSums[row1-1][col2];
else
return matrixSums[row2][col2] + matrixSums[row1-1][col1-1] - matrixSums[row2][col1-1] - matrixSums[row1-1][col2];
}
}