304. Range Sum Query 2D

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.

1. You may assume that row1 ≤ row2 and col1 ≤ col2.

了解积分图的同学可能一眼就看出来这个问题怎么解决了。

下面贴代码。讨论区也有一个大神给出了很好的解释。

class NumMatrix {private:    int row, col;    vector<vector<int>> sums;public:    NumMatrix(vector<vector<int>> &matrix) {        row = matrix.size();        col = row>0 ? matrix[0].size() : 0;        sums = vector<vector<int>>(row+1, vector<int>(col+1, 0));        for(int i=1; i<=row; i++) {            for(int j=1; j<=col; j++) {                sums[i][j] = matrix[i-1][j-1] +                              sums[i-1][j] + sums[i][j-1] - sums[i-1][j-1] ;            }        }    }    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];    }};