304. Range Sum Query 2D - Immutable

Posted 2021-03-25 wentiliangkaihua

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.

class NumMatrix {

    private int[][] dp;
    public NumMatrix(int[][] matrix) {
        if(matrix.length == 0 || matrix[0].length == 0) return;//事实上起作用的是前半句，后半句可以不要
        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）