标签：span 题目 for color etc osi style else block
题目如下：
Given a
m * nmatrixmatand an integerK, return a matrixanswerwhere eachanswer[i][j]is the sum of all elementsmat[r][c]fori - 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.lengthn == mat[i].length1 <= m, n, K <= 1001 <= mat[i][j] <= 100
解题思路：记grid[i][j] 为 左上角顶点是(0,0)，右下角顶点是(i,j) 的矩阵和，grid可以通过时间复杂度为O(n^2)的循环计算出来，接下来再遍历mat，计算出相应每个子矩阵和即可。
代码如下：
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
【leetcode】1314. Matrix Block Sum
标签：span 题目 for color etc osi style else block
原文地址：https://www.cnblogs.com/seyjs/p/12183197.html
