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:
- 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.
这道题让我们求一个二维区域和的检索,是之前那道题Range Sum Query - Immutable 区域和检索的延伸。有了之前那道题的基础,我们知道这道题其实也是换汤不换药,还是要建立一个累计区域和的数组,然后根据边界值的加减法来快速求出给定区域之和。这里我们维护一个二维数组dp,其中dp[i][j]表示累计区间(0, 0)到(i, j)这个矩形区间所有的数字之和,那么此时如果我们想要快速求出(r1, c1)到(r2, c2)的矩形区间时,只需dp[r2][c2] - dp[r2][c1 - 1] - dp[r1 - 1][c2] + dp[r1 - 1][c1 - 1]即可,下面的代码中我们由于用了辅助列和辅助行,所以下标会有些变化,参见代码如下:
class NumMatrix { public: NumMatrix(vector<vector<int> > &matrix) { if (matrix.empty() || matrix[0].empty()) return; dp.resize(matrix.size() + 1, vector<int>(matrix[0].size() + 1, 0)); for (int i = 1; i <= matrix.size(); ++i) { for (int j = 1; j <= matrix[0].size(); ++j) { dp[i][j] = dp[i][j - 1] + dp[i - 1][j] - dp[i - 1][j - 1] + matrix[i - 1][j - 1]; } } } int sumRegion(int row1, int col1, int row2, int col2) { return dp[row2 + 1][col2 + 1] - dp[row1][col2 + 1] - dp[row2 + 1][col1] + dp[row1][col1]; } private: vector<vector<int> > dp; };
本文转自博客园Grandyang的博客,原文链接:二维区域和检索 - 不可变[LeetCode] Range Sum Query 2D - Immutable ,如需转载请自行联系原博主。