LeetCode 304 Range Sum Query 2D - Immutable

LeetCode

Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (_row_1, _col_1) and lower right corner (_row_2, _col_2).

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 _row_1 ≤ _row_2 and _col_1 ≤ _col_2.

分析

和 303 题一样，只不过变成二维数组。那么就分别把每一行用 303 的解法。

代码

public class NumMatrix {
    public long[][] sumMatrix;
    public NumMatrix(int[][] matrix) {
        if (matrix == null || matrix.length == 0) {
            return;
        }
        sumMatrix = new long[matrix.length + 1][matrix[0].length + 1];
        for (int i = 0; i < matrix.length; i++) {
            for (int j = 0; j < matrix[0].length; j++) {
                sumMatrix[i][j + 1] = sumMatrix[i][j] + matrix[i][j];
            }
        }
    }
    public int sumRegion(int row1, int col1, int row2, int col2) {
        if (sumMatrix == null || row1 < 0 || row2 < 0 || col1 < 0
                || col2 < 0 || row1 >= sumMatrix.length - 1
                || row2 >= sumMatrix.length - 1
                || col1 >= sumMatrix[0].length - 1
                || col2 >= sumMatrix[0].length - 1 || row1 > row2
                || col1 > col2) {
            return Integer.MIN_VALUE;
        }
        long rt = 0;
        for (int i = row1; i <= row2; i++) {
            rt += sumMatrix[i][col2 + 1] - sumMatrix[i][col1];
        }
        return (int) rt;
    }
}
// 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);