Leetcode 3619. Count Islands With Total Value Divisible by K

• Leetcode 3619. Count Islands With Total Value Divisible by K
  • 1. 解题思路
  • 2. 代码实现

• 题目链接：3619. Count Islands With Total Value Divisible by K

1. 解题思路

这一题思路上就是一个图的遍历，我们首先顺序访问每一个格点，如果某一个格点所属的岛还没有被访问过，那么就对其进行一下深度优先遍历，将岛上所有格点进行标记，然后返回整个岛的value，考察其是否满足条件。

然后，我们返回所有符合条件的岛的个数即可。

2. 代码实现

给出python代码实现如下：

class Solution:
    def countIslands(self, grid: List[List[int]], k: int) -> int:
        n, m = len(grid), len(grid[0])
        status = [[0 for _ in range(m)] for _ in range(n)]

        def dfs(i, j):
            nonlocal status
            s = [(i, j)]
            seen = {(i, j)}
            ans = 0
            while s:
                i, j = s.pop(0)
                ans += grid[i][j]
                status[i][j] = 1
                if i-1 >= 0 and grid[i-1][j] != 0 and status[i-1][j] == 0 and (i-1, j) not in seen:
                    seen.add((i-1, j))
                    s.append((i-1, j))
                if i+1 < n and grid[i+1][j] != 0 and status[i+1][j] == 0 and (i+1, j) not in seen:
                    seen.add((i+1, j))
                    s.append((i+1, j))
                if j-1 >= 0 and grid[i][j-1] != 0 and status[i][j-1] == 0 and (i, j-1) not in seen:
                    seen.add((i, j-1))
                    s.append((i, j-1))
                if j+1 < m and grid[i][j+1] != 0 and status[i][j+1] == 0 and (i, j+1) not in seen:
                    seen.add((i, j+1))
                    s.append((i, j+1))
            return ans

        ans = 0
        for i in range(n):
            for j in range(m):
                if grid[i][j] != 0 and status[i][j] == 0:
                    val = dfs(i, j)
                    if val % k == 0:
                        ans += 1
        return ans

提交代码评测得到：耗时311ms，占用内存37.94MB。