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跟上一題異曲同工，有了上個動態規劃的思想，擴充到這個題，就是：
每個存它到矩陣左上角的和，然後算的時候就用右下角減上方的一個，減左方的一個，再加左上角的一個。
代碼如下：
class NumMatrix {
public:
NumMatrix(vector<vector<int>> matrix) {
for(int i = 0; i < matrix.size(); ++i){
vector<int> temp;
for(int j = 0; j < matrix[i].size(); ++j){
temp.push_back(0);
}
matrix.push_back(temp);
}
for(int i = 0; i < matrix.size(); ++i){
for(int j = 0; j < matrix[i].size(); ++j){
for(int m = 0; m <= i; ++m){
for(int n = 0; n <= j; ++n){
summm[i][j]+=matrix[n][n];
}
}
}
}
}
int sumRegion(int row1, int col1, int row2, int col2) {
if(row1 != 0) row1 = row1 -1;
if(col1 != 0) col1 = col1 -1;
return summm[row2][col2]+summm[row1][col1]-summm[row2][col1]-summm[row1][col2];
}
vector<vector<int>>summm;
};
/**
* Your NumMatrix object will be instantiated and called as such:
* NumMatrix obj = new NumMatrix(matrix);
* int param_1 = obj.sumRegion(row1,col1,row2,col2);
*/