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:
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.
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给定一个矩阵，求其中任意一个子矩阵中值的和。题目说了求和的要求比较多，所以要尽量使得求和的复杂度小一点。在构造函数里就建立一个矩阵，记录从(0,0)到(i,j)子矩阵的和。然后求和的时候,，可以用从(0,0)到(row2,col2)的子矩阵减去从(0,0)到(row2,col1-1)的子矩阵，再减去从(0,0)到(row1-1,col2)的子矩阵，最后再加上从(0,0)到(row1-1,col1-1)的子矩阵（因为这部分区域减了两次）得到答案。
代码：
class NumMatrix { public: NumMatrix(vector<vector<int> > matrix) { if(matrix.empty()) return; sum.assign(matrix.size()+1, vector<int>(matrix[0].size()+1, 0)); int rowsum = 0; for(int i = 0; i < matrix.size(); ++i) { for(int j = 0; j < matrix[i].size(); ++j) { rowsum += matrix[i][j]; sum[i+1][j+1] = sum[i][j+1] + rowsum; } rowsum = 0; } } int sumRegion(int row1, int col1, int row2, int col2) { return sum[row2+1][col2+1]- sum[row1][col2+1] - sum[row2+1][col1] + sum[row1][col1]; } private: vector<vector<int> > sum; };
