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：

 class NumMatrix {
 public:
     NumMatrix(vector<vector<int>> &matrix){
         if(!matrix.size() || !matrix[].size())
             return;
         sum = vector<vector<int>>(matrix.size(), vector<int>(matrix[].size(), ));
         for(int i = ; i < matrix.size(); ++i){
             for(int j = ; j < matrix[].size(); ++j){
                 if(i !=  && j != ){
                     sum[i][j] = matrix[i][j] + sum[i][j-] + sum[i-][j] - sum[i-][j-];
                 }else if(i ==  && j == ){
                     sum[i][j] = matrix[i][j];
                 }else if(i == ){
                     sum[i][j] = matrix[i][j] + sum[i][j-];
                 }else{
                     sum[i][j] = matrix[i][j] + sum[i-][j];
                 }
             }
         }
     }

     int sumRegion(int row1, int col1, int row2, int col2) {
         if(row1 ==  && col1 == )
             return sum[row2][col2];
         else if(row1 == )
             return sum[row2][col2] - sum[row2][col1-];
         else if(col1 == )
             return sum[row2][col2] - sum[row1-][col2];
         else
             return sum[row2][col2] - sum[row2][col1-] -  sum[row1-][col2] + sum[row1-][col1-];

     }
 private:
     vector<vector<int>> sum;
 };