Range Sum Query 2D -- leetcode

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.

建立一个映射：以(0,0)点为左上角，以当前元素为右下角，围着的一个矩形。矩形中所有元素的和为一个值sum。

映射：f(x,y) --> sum(0,0, x,y)

则：

sum(x1,y1, x2, y2) = f(x2,y2) - f(x2-1,y2) - f(x2, y2-1) + f(x2-1, y2-1)

为了免去判断越界，即，x, y是否为0。

可以在最左侧，最上侧，全填充0.

而对应的坐标也进行变化。 x_n = x_o + 1;  y_n = y_o + 1

求sum(x1,y1, x2, y2) 则变成；

sum(x1+1, y1+1, x2+1, y2+1) = f(x2+1, y2+1) - f(x2, y2+1) - f(x2+1, y2) + f(x2, y2)

class NumMatrix {    vector<vector<int> > sums;public:    NumMatrix(vector<vector<int>> &matrix) {        const int m = matrix.size();        const int n = matrix.empty() ? 0 : matrix[0].size();        sums = vector<vector<int> >(m+1, vector<int>(n+1));        for (int i=0; i<m; i++) {            for (int j=0; j<n; j++) {                sums[i+1][j+1] = sums[i][j+1] + sums[i+1][j] - sums[i][j] + matrix[i][j];            }        }    }    int sumRegion(int row1, int col1, int row2, int col2) {        return sums[row2+1][col2+1] - sums[row2+1][col1] - sums[row1][col2+1] + sums[row1][col1];    }};// Your NumMatrix object will be instantiated and called as such:// NumMatrix numMatrix(matrix);// numMatrix.sumRegion(0, 1, 2, 3);// numMatrix.sumRegion(1, 2, 3, 4);