LeetCode304

Range Sum Query 2D - Immutable

題目：

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.

題目大意：給一個二維矩陣，求出左上角x1,y1到右下角x2,y2座標內所有數的和。

題目有提示，這個方法會被調用很多次，普通方法超時不多說。。。所以用動態規劃。

一個二維數組sum[r][c]表示r行，c列到0行，0列的數的和。

則sum[r,c] = matrix[r,c]+sum[r-1,c]+sum[r,c-1]-sum[r-1,c-1];

求sumRegion時

sumRegion(row1,col1,row2,col2) =
sum[row1-1,col1-1]+sum[row2,col2]-sum[row1-1,col2]-sum[row2,col1-1]

所以完整代碼如下：

public class NumMatrix {
    int[,] sum;
    public NumMatrix(int[,] matrix)
    {
        if (matrix.GetLength(0) == 0 || matrix.GetLength(1) == 0 || matrix == null) return;
        sum = new int[matrix.GetLength(0),matrix.GetLength(1)];
        sum[0,0] = 0;
        for (int r = 0; r < matrix.GetLength(0); r++)
        {
            for (int c = 0; c < matrix.GetLength(1); c++)
            {
                if (r == 0 && c == 0) sum[r, c] = matrix[r, c];
                else if (r == 0) sum[r,c]=matrix[r,c]+sum[r,c-1];
                else if (c == 0) sum[r,c] = matrix[r,c] + sum[r-1,c];
                else sum[r,c] = matrix[r,c]+sum[r-1,c]+sum[r,c-1]-sum[r-1,c-1];
            }
        }
    }

    public int SumRegion(int row1, int col1, int row2, int col2)
    {
        if (row1 == 0 && col1 == 0) return sum[row2, col2];
        if (row1 == 0) return sum[row2, col2] - sum[row2, col1 - 1];
        if (col1 == 0) return sum[row2,col2] - sum[row1-1,col2];
        return sum[row1-1,col1-1]+sum[row2,col2]-sum[row1-1,col2]-sum[row2,col1-1];
    }
}