算法: 求矩形框数字之和304. Range Sum Query 2D - Immutable

Given a 2D matrix matrix, handle multiple queries of the following type:

• Calculate the sum of the elements of matrix inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).

Implement the NumMatrix class:

• NumMatrix(int[][] matrix) Initializes the object with the integer matrix matrix.
• int sumRegion(int row1, int col1, int row2, int col2) Returns the sum of the elements of matrix inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).

Example 1:

Input
["NumMatrix", "sumRegion", "sumRegion", "sumRegion"]
[[[[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]]], [2, 1, 4, 3], [1, 1, 2, 2], [1, 2, 2, 4]]
Output
[null, 8, 11, 12]

Explanation
NumMatrix numMatrix = new NumMatrix([[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]]);
numMatrix.sumRegion(2, 1, 4, 3); // return 8 (i.e sum of the red rectangle)
numMatrix.sumRegion(1, 1, 2, 2); // return 11 (i.e sum of the green rectangle)
numMatrix.sumRegion(1, 2, 2, 4); // return 12 (i.e sum of the blue rectangle)

Constraints:

• m == matrix.length
• n == matrix[i].length
• 1 <= m, n <= 200
• -105 <= matrix[i][j] <= 105
• 0 <= row1 <= row2 < m
• 0 <= col1 <= col2 < n
• At most 10^4 calls will be made to sumRegion.

动态规划解法

class NumMatrix {
    int[][] sum;
    public NumMatrix(int[][] matrix) {
        int rows = matrix.length;
        int cols = matrix[0].length;
        sum = new int[rows + 1][cols + 1];// sum[i][j] is sum of all elements inside the rectangle [0,0,i,j]
        for (int r = 1; r <= rows; r++) {
            for (int c = 1; c <= cols; c++) {
                sum[r][c] = sum[r - 1][c] + sum[r][c - 1] + matrix[r - 1][c - 1] - sum[r - 1][c - 1];
            }
        }
    }
    
    public int sumRegion(int row1, int col1, int row2, int col2) {
     	// Since our `sum` starts by 1 so we need to increase r1, c1, r2, c2 by 1
        return sum[row2 + 1][col2 + 1] - sum[row1][col2 + 1] - sum[row2 + 1][col1] + sum[row1][col1];
    }
}

/**
 * Your NumMatrix object will be instantiated and called as such:
 * NumMatrix obj = new NumMatrix(matrix);
 * int param_1 = obj.sumRegion(row1,col1,row2,col2);
 */

或者参考视频：https://youtu.be/PwDqpOMwg6U

参考

https://leetcode.com/problems/range-sum-query-2d-immutable/discuss/572648/C%2B%2BJavaPython-Prefix-sum-with-Picture-explain-Clean-and-Concise

https://youtu.be/PwDqpOMwg6U