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.
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.
题目和前面#303题相同，只不过将一维数组换成了二维矩阵。用与前一题类似的方法，我们用mymatrix[i][j]表示矩阵中(i+1,j+1)到(0,0)元素中所有元素的和，如图所示： 蓝色的点为matrix[i][j]元素，两个橘色的点为matrix[i - 1][j]和matrix[i][j - 1]元素，绿点为matrix[i - 1][j - 1]元素，由图可见，如果我们计算matrix[i][j] + mymatrix[i][j + 1] + mymatrix[i + 1][j]，阴影部分会被叠加两次，应减掉，因此我们计算mymatrix为 mymatrix[i + 1][j + 1] = matrix[i][j] + mymatrix[i][j + 1] + mymatrix[i + 1][j] - mymatrix[i][j]。 下面就使用mymatrix数组去计算要求的sumRegion了。继续看图： 蓝色的点为(row2, col2)元素，绿色的点为(row1, col1)元素。我们要求的sumRegion就是图中黄色阴影部分。如果我们计算mymatrix[row2 + 1][col2 + 1] - mymatrix[row2 + 1][col1] - mymatrix[row1][col2 + 1]，即图中绿色矩形减去两个红色矩形，那么蓝色阴影部分被减了两次，应加回。因此我们计算最后的sumRegion为： mymatrix[row2 + 1][col2 + 1] - mymatrix[row2 + 1][col1] - mymatrix[row1][col2 + 1] + mymatrix[row1][col1]; 具体实现代码如下所示。代码的空间复杂度为 O(n2) ，创建NumMatrix的时间复杂度为 O(n2) ，计算sumRegion的时间复杂度为 O(1) 。
class NumMatrix { public: NumMatrix(vector<vector<int>> matrix){ int rows = matrix.size(); int cols = rows == 0 ? 0 : matrix[0].size(); mymatrix = vector<vector<int>>(rows+1, vector<int>(cols+1, 0)); for(int i = 0; i < rows; i++){ for(int j = 0; j < cols; j++){ mymatrix[i + 1][j + 1] = matrix[i][j] + mymatrix[i][j + 1] + mymatrix[i + 1][j] - mymatrix[i][j]; } } } int sumRegion(int row1, int col1, int row2, int col2) { return mymatrix[row2 + 1][col2 + 1] - mymatrix[row2 + 1][col1] - mymatrix[row1][col2 + 1] + mymatrix[row1][col1]; } private: vector<vector<int>> mymatrix; }; /** * Your NumMatrix object will be instantiated and called as such: * NumMatrix obj = new NumMatrix(matrix); * int param_1 = obj.sumRegion(row1,col1,row2,col2); */