304 Range Sum Query 2D - immutable

Given a 2D matrixmatrix, 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.

这题就是求二维的prefix sum。求prefix Sum的方法是，左下+右上+这个值 - 左上。查询时是：右下的（i2+1，j2+1）- （i1，j2 +1）- （i2 + 1， j1）+ (i1, j1)

int[][] preSumM;
public NumMatrix(int[][] matrix) {
    if (matrix.length == 0 || matrix[0].length == 0) {
        return;
    }

    int n = matrix.length;
    int m = matrix[0].length;

    preSumM = new int[n + 1][m + 1];
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            preSumM[i][j] = preSumM[i - 1][j] + preSumM[i][j - 1] + matrix[i - 1][j - 1] - preSumM[i - 1][j - 1];
        }
    }
}

public int sumRegion(int row1, int col1, int row2, int col2) {
    return preSumM[row2 + 1][col2 + 1] + preSumM[row1][col1] - preSumM[row1][col2 + 1] - preSumM[row2 + 1][col1];
}