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) -> 8 sumRegion(1, 1, 2, 2) -> 11 sumRegion(1, 2, 2, 4) -> 12跟上一题异曲同工，有了上个动态规划的思想，扩充到这个题，就是：
每个存它到矩阵左上角的和，然后算的时候就用右下角减上方的一个，减左方的一个，再加左上角的一个。
代码如下：
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);
*/