我报名参加金石计划1期挑战——瓜分10万奖池，这是我的第1篇文章，点击查看活动详情
描述
You are given an n x n integer matrix grid. Generate an integer matrix maxLocal of size (n - 2) x (n - 2) such that:
- maxLocal[i][j] is equal to the largest value of the 3 x 3 matrix in grid centered around row i + 1 and column j + 1.
In other words, we want to find the largest value in every contiguous 3 x 3 matrix in grid. Return the generated matrix.
Example 1:
Input: grid = [[9,9,8,1],[5,6,2,6],[8,2,6,4],[6,2,2,2]]
Output: [[9,9],[8,6]]
Explanation: The diagram above shows the original matrix and the generated matrix.
Notice that each value in the generated matrix corresponds to the largest value of a contiguous 3 x 3 matrix in grid.
Example 2:
Input: grid = [[1,1,1,1,1],[1,1,1,1,1],[1,1,2,1,1],[1,1,1,1,1],[1,1,1,1,1]]
Output: [[2,2,2],[2,2,2],[2,2,2]]
Explanation: Notice that the 2 is contained within every contiguous 3 x 3 matrix in grid.
Note:
n == grid.length == grid[i].length
3 <= n <= 100
1 <= grid[i][j] <= 100
解析
根据题意，给定一个 n x n 整数矩阵 grid 。 生成一个大小为 (n - 2) x (n - 2) 的整数矩阵 maxLocal ，使得：
- maxLocal[i][j] 等于以 i + 1 行和 j + 1 列为中心的网格中 3 x 3 矩阵的最大值。
换句话说，我们想在网格中的每个连续 3 x 3 矩阵中找到最大值。 返回生成的矩阵。
这道题其实就是考察卷积神经网络中的池化，我们只需要对不同的九宫格取最大值，然后将这些最大值组合起来就是最后的结果矩阵。
时间复杂度为 O(N^2 * 9) ，空间复杂度为 O(N^2) 。
解答
class Solution(object):
def largestLocal(self, grid):
"""
:type grid: List[List[int]]
:rtype: List[List[int]]
"""
N = len(grid)
result = [[0] * (N-2) for _ in range(N-2)]
for i in range(1, N-1):
for j in range(1, N-1):
result[i-1][j-1] = max(grid[i-1][j-1],grid[i-1][j],grid[i-1][j+1],
grid[i][j-1],grid[i][j],grid[i][j+1],
grid[i+1][j-1],grid[i+1][j],grid[i+1][j+1])
return result
运行结果
50 / 50 test cases passed.
Status: Accepted
Runtime: 100 ms
Memory Usage: 14.1 MB
原题链接
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