LeetCode OJ：Range Sum Query 2D - Immutable（区域和2D版本）

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).

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

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.

前面一个博文的衍生版本，由原来的一维矩阵变成了现在的二维矩阵，没什么区别，还是先建立和的数组，代码如下所示：

 1 class NumMatrix {
 2 public:
 3     NumMatrix(vector<vector<int>> &matrix){
 4         if(!matrix.size() || !matrix[0].size())
 5             return;
 6         sum = vector<vector<int>>(matrix.size(), vector<int>(matrix[0].size(), 0));
 7         for(int i = 0; i < matrix.size(); ++i){
 8             for(int j = 0; j < matrix[0].size(); ++j){
 9                 if(i != 0 && j != 0){
10                     sum[i][j] = matrix[i][j] + sum[i][j-1] + sum[i-1][j] - sum[i-1][j-1];
11                 }else if(i == 0 && j == 0){
12                     sum[i][j] = matrix[i][j];
13                 }else if(i == 0){
14                     sum[i][j] = matrix[i][j] + sum[i][j-1];
15                 }else{
16                     sum[i][j] = matrix[i][j] + sum[i-1][j];
17                 }
18             }
19         }
20     }
21
22     int sumRegion(int row1, int col1, int row2, int col2) {
23         if(row1 == 0 && col1 == 0)
24             return sum[row2][col2];
25         else if(row1 == 0)
26             return sum[row2][col2] - sum[row2][col1-1];
27         else if(col1 == 0)
28             return sum[row2][col2] - sum[row1-1][col2];
29         else
30             return sum[row2][col2] - sum[row2][col1-1] -  sum[row1-1][col2] + sum[row1-1][col1-1];
31
32     }
33 private:
34     vector<vector<int>> sum;
35 };