Range Sum Query 2D - Immutable

题目描述：

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.

这个题和 Range Sum Query - Immutable一个样，用newMatrix[i][j]来存储(0,0)到(i,j)的矩形中的数的综合，newMatrix[i][j]=newMatrix[i][j-1]+newMatrix[i-1][j]-newMatrix[i-1][j-1]+matrix[i][j];然后再求area[i][j]。

代码如下：

public class NumMatrix {
    int[][] newMatrix;
	public NumMatrix(int[][] matrix) {
	    //注意0,0
	    if(matrix.length==0||matrix[0].length==0)
	        return;
        newMatrix=new int[matrix.length][matrix[0].length];
        newMatrix[0][0]=matrix[0][0];
        for(int j=1;j<matrix[0].length;j++){
        	newMatrix[0][j]=newMatrix[0][j-1]+matrix[0][j];
        }
        for(int i=1;i<matrix.length;i++){
        	newMatrix[i][0]=newMatrix[i-1][0]+matrix[i][0];
        }
        for(int i=1;i<matrix.length;i++){
        	for(int j=1;j<matrix[0].length;j++){
        		newMatrix[i][j]=newMatrix[i][j-1]+newMatrix[i-1][j]-newMatrix[i-1][j-1]+matrix[i][j];
        	}
        }
    }

    public int sumRegion(int row1, int col1, int row2, int col2) {
    	int area1=row1==0?0:newMatrix[row1-1][col2];
    	int area2=col1==0?0:newMatrix[row2][col1-1];
    	int area3=(row1==0||col1==0)?0:newMatrix[row1-1][col1-1];
        return newMatrix[row2][col2]-area1-area2+area3;
    }
}