LeetCode Q304: 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.

Solution:

Use sum region. My answer takes 24ms in C++ and beat 100% of all submissions.

	class NumMatrix {
	public:
	NumMatrix(vector<vector<int>> &matrix) {
	if(matrix.empty())
	return;
	mat.resize(matrix.size()+1);
	for(int i=0; i<mat.size(); i++){
	mat[i].resize(matrix[0].size()+1);
	}
	for(int i=1; i<=matrix.size(); i++){
	for(int j=1; j<=matrix[0].size(); j++){
	mat[i][j] = mat[i][j-1]+mat[i-1][j]-mat[i-1][j-1]+matrix[i-1][j-1];
	}
	}
	}
	int sumRegion(int row1, int col1, int row2, int col2) {
	if(mat.empty())
	return 0;
	int sum = mat[row2+1][col2+1] - mat[row2+1][col1]-mat[row1][col2+1]+mat[row1][col1];
	return sum;
	}
	vector<vector<int>> mat;
	};
	// Your NumMatrix object will be instantiated and called as such:
	// NumMatrix numMatrix(matrix);
	// numMatrix.sumRegion(0, 1, 2, 3);
	// numMatrix.sumRegion(1, 2, 3, 4);

view raw Q304.cpp hosted with ❤ by GitHub