leetcode(29)矩阵搜索系列题目

64. 最小路径和

动态规划

class Solution:
    def minPathSum(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        res = 0
        dp = [[0]* n for _ in range(m)]  #注意先列再行
        dp[0][0] = grid[0][0]
        for i in range(1, m):
            dp[i][0] = dp[i - 1][0] + grid[i][0]
        for j in range(1, n):
            dp[0][j] = dp[0][j - 1] + grid[0][j]
        for i in range(1, m):
            for j in range(1, n):
                dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j]
        return dp[-1][-1]

6148. 矩阵中的局部最大值

本质上是在做最大池化

class Solution:
    def largestLocal(self, grid: List[List[int]]) -> List[List[int]]:
        n = len(grid[0])
        res = [[1]*(n - 2) for _ in range(n-2)] 
        for i in range(n-2):
            for j in range(n-2):
                max_num = 1
                for p in range(i, i+3):
                    for q in range(j, j+3):
                        max_num = max(max_num, grid[p][q])
                res[i][j] = max_num
        return res

另一种优化的方法，原地修改，不需要使用一个新矩阵来返回

class Solution:
    def largestLocal(self, grid: List[List[int]]) -> List[List[int]]:
        n = len(grid[0])
        for i in range(n-2):
            for j in range(n-2):
                grid[i][j] = max(max(raw[j:j+3]) for raw in grid[i:i+3])
            grid[i].pop()  # 删除前面两列
            grid[i].pop()
        grid.pop()  # 删除前面两行
        grid.pop()
        return grid