Range Sum Query 2D – Immutable
Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by (row1, col1), (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) -> 12Note:
- You may assume that the matrix does not change.
- There are many calls to sumRegion function.
- You may assume that row1 ≤ row2 and col1 ≤ col2.
Dynamic programming.
dp[i][j] stores the sum of elements in the rectangle area between(0,0) and (i,j)
class NumMatrix { public: vector<vector<long>> dp; NumMatrix(vector<vector<int>> &matrix) { if(matrix.size() == 0 || matrix[0].size() == 0) return ; dp = vector<vector<long>> (matrix.size(), vector<long>(matrix[0].size(), 0)); dp[0][0] = matrix[0][0]; for(int i = 0; i < matrix.size(); i++){ for(int j = 0; j < matrix[0].size(); j++){ if(j == 0 && i == 0) continue; else if(j == 0){ dp[i][j] = dp[i - 1][j] + matrix[i][j]; } else if(i == 0){ dp[i][j] = dp[i][j - 1] + matrix[i][j]; } else{ dp[i][j] = dp[i][j - 1] + dp[i - 1][j] - dp[i - 1][j - 1] + matrix[i][j]; } } } } int sumRegion(int row1, int col1, int row2, int col2) { int sum = 0; // index boundary check if(row1 == 0 && col1 == 0) sum = (int)dp[row2][col2]; else if(row1 == 0){ sum = (int)(dp[row2][col2] - dp[row2][col1 - 1]); }else if(col1 == 0){ sum = (int)(dp[row2][col2] - dp[row1 - 1][col2]); }else{ sum = (int)(dp[row2][col2] + dp[row1 - 1][col1 - 1] - dp[row2][col1 - 1] - dp[row1 - 1][col2]); } return sum; } }; // 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);