304. Range Sum Query 2D - Immutable

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).

304. Range Sum Query 2D - Immutable
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.
class NumMatrix {

    private int[][] dp;
    public NumMatrix(int[][] matrix) {
        if(matrix.length == 0 || matrix[0].length == 0) return;
        this.dp = new int[matrix.length][matrix[0].length];
        for(int i = 0; i < matrix.length; i++){
            int sum = 0;
            for(int j = 0; j < matrix[0].length; j++){
                sum += matrix[i][j];
                dp[i][j] = sum;
            }
        }
    }
    
    public int sumRegion(int row1, int col1, int row2, int col2) {
        int res = 0;
        for(int i = row1; i <= row2; i++){
            res += (dp[i][col2] - (col1 == 0 ? 0 : dp[i][col1 - 1]));
        }
        return res;
    }
}

二维和一维无本质区别,把每行看作一个一维即可。注意判断空数组的情况(直接return 为null)

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