LeetCode304

Range Sum Query 2D - Immutable

题目：

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) -> 8sumRegion(1, 1, 2, 2) -> 11sumRegion(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.

题目大意：给一个二维矩阵，求出左上角x1,y1到右下角x2,y2坐标内所有数的和。

题目有提示，这个方法会被调用很多次，普通方法超时不多说。。。所以用动态规划。

一个二维数组sum[r][c]表示r行，c列到0行，0列的数的和。

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

求sumRegion时

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

所以完整代码如下：

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];    }}