剑指 Offer II 013. 二维子矩阵的和
给定一个二维矩阵 matrix,以下类型的多个请求:
计算其子矩形范围内元素的总和,该子矩阵的左上角为 (row1, col1) ,右下角为 (row2, col2) 。
实现 NumMatrix 类:
NumMatrix(int[][] matrix) 给定整数矩阵 matrix 进行初始化
int sumRegion(int row1, int col1, int row2, int col2) 返回左上角 (row1, col1) 、右下角 (row2, col2) 的子矩阵的元素总和。
示例 1:
输入:
["NumMatrix","sumRegion","sumRegion","sumRegion"]
[[[[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]]],[2,1,4,3],[1,1,2,2],[1,2,2,4]]
输出:
[null, 8, 11, 12]
解释:
NumMatrix numMatrix = new NumMatrix([[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]]]);
numMatrix.sumRegion(2, 1, 4, 3); // return 8 (红色矩形框的元素总和)
numMatrix.sumRegion(1, 1, 2, 2); // return 11 (绿色矩形框的元素总和)
numMatrix.sumRegion(1, 2, 2, 4); // return 12 (蓝色矩形框的元素总和)
提示:
m == matrix.length
n == matrix[i].length
1 <= m, n <= 200
-105 <= matrix[i][j] <= 105
0 <= row1 <= row2 < m
0 <= col1 <= col2 < n
最多调用 104 次 sumRegion 方法
解析:
对二维矩阵求一个前缀和,每次query减去上边的和左边的即可,注意有重叠
class NumMatrix { public: vector<vector<int>> matrix; int presum[210][210]; int n, m; NumMatrix(vector<vector<int>>& matrix) { this->matrix = matrix; int n = matrix.size(); int m = matrix[0].size(); this->n = n; this->m = m; for(int i = 0; i < n; i++) { int rsum = 0; for(int j = 0; j < m; j++) { if(i == 0 && j == 0) presum[i][j] = matrix[i][j]; else if(i == 0) presum[i][j] = presum[i][j - 1] + matrix[i][j]; else { rsum += matrix[i][j]; presum[i][j] = presum[i - 1][j] + rsum; } } } } int sumRegion(int row1, int col1, int row2, int col2) { int temp =presum[row2][col2]; int temp1 = 0, temp2 = 0; if(row1 - 1 >= 0) { temp1 = presum[row1 - 1][col2]; } if(col1 - 1 >= 0) { if(row1 - 1 >= 0) temp2 = presum[row2][col1 - 1] - presum[row1 - 1][col1 - 1]; else temp2 = presum[row2][col1 - 1]; } return temp - temp1 - temp2; } }; /** * Your NumMatrix object will be instantiated and called as such: * NumMatrix* obj = new NumMatrix(matrix); * int param_1 = obj->sumRegion(row1,col1,row2,col2); */