【leetcode】1314. Matrix Block Sum

題目以下：

Given a m * n matrix mat and an integer K, return a matrix answer where each answer[i][j] is the sum of all elements mat[r][c] for i - K <= r <= i + K, j - K <= c <= j + K, and (r, c) is a valid position in the matrix.

Example 1:

Input: mat = [[1,2,3],[4,5,6],[7,8,9]], K = 1
Output: [[12,21,16],[27,45,33],[24,39,28]]

Example 2:

Input: mat = [[1,2,3],[4,5,6],[7,8,9]], K = 2
Output: [[45,45,45],[45,45,45],[45,45,45]]

Constraints:

• m == mat.length
• n == mat[i].length
• 1 <= m, n, K <= 100
• 1 <= mat[i][j] <= 100

解題思路：記grid[i][j] 爲 左上角頂點是(0,0)，右下角頂點是(i,j) 的矩陣和，grid能夠經過時間複雜度爲O(n^2)的循環計算出來，接下來再遍歷mat，計算出相應每一個子矩陣和便可。

代碼以下：

class Solution(object): def matrixBlockSum(self, mat, K): """ :type mat: List[List[int]] :type K: int :rtype: List[List[int]] """ res = [[0] * len(mat[0]) for _ in mat] grid = [[0] * len(mat[0]) for _ in mat] for i in range(len(mat)): count = 0 for j in range(len(mat[0])): count += mat[i][j] grid[i][j] = count for i in range(len(res)): for j in range(len(res[i])): left = max(0,j-K) right = min(j+K,len(res[i])-1) count = 0 for row in range(max(0,i-K),min(i+K+1,len(res))): if left == 0: count += grid[row][right] else: count += (grid[row][right] - grid[row][left-1]) res[i][j] = count return res