[leetcode-304-Range Sum Query 2D - Immutable]

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

 

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.

思路：

Construct a 2D array sums[row+1][col+1]

(notice: we add additional blank row sums[0][col+1]={0} and blank column sums[row+1][0]={0} to remove the edge case checking), so, we can have the following definition

sums[i+1][j+1] represents the sum of area from matrix[0][0] to matrix[i][j]

To calculate sums, the ideas as below

And we can have the following code

vector<vector<int>> sum;
  NumMatrix(vector<vector<int>> matrix)
{         
      int row = matrix.size(),col = matrix[0].size();
      sum = vector<vector<int>>(row+1,vector<int>(col+1,0)) ;
      
      for(int i=1;i<=row;i++)
      {
        for(int j =1;j<=col;j++)
        {
          sum[i][j] = matrix[i-1][j-1] + sum[i][j-1] + sum[i-1][j] - sum[i-1][j-1]; 
        }
      }    
}
    
    int sumRegion(int row1, int col1, int row2, int col2) 
    {
        return sum[row2+1][col2+1] - sum[row2+1][col1] - sum[row1][col2+1] + sum[row1][col1];
    }

 参考：

https://discuss.leetcode.com/topic/29536/clean-c-solution-and-explaination-o-mn-space-with-o-1-time