[leetcode] 694. Number of Distinct Islands @ python_leetcode 694 python-优快云博客

本文链接：https://blog.youkuaiyun.com/danspace1/article/details/86610850

本文介绍了一种使用深度优先搜索(DFS)算法来解决岛屿计数问题的方法。在给定一个由0和1组成的二维数组中，1表示陆地，0表示水。目标是计算不重复岛屿的数量，其中岛屿是通过水平或垂直相连的陆地块组成，且不允许旋转或反射。通过定义一个dfs函数，递归计算岛屿相对于左上角的偏移量，将偏移量转化为元组并存入集合中，最终返回集合的大小即为不同岛屿的数量。

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原题

Given a non-empty 2D array grid of 0’s and 1’s, an island is a group of 1’s (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

Count the number of distinct islands. An island is considered to be the same as another if and only if one island can be translated (and not rotated or reflected) to equal the other.

Example 1:
11000
11000
00011
00011
Given the above grid map, return 1.
Example 2:
11011
10000
00001
11011
Given the above grid map, return 3.

Notice that:
11
1
and
1
11
are considered different island shapes, because we do not consider reflection / rotation.
Note: The length of each dimension in the given grid does not exceed 50.

解法

参考: Simple Python 169ms
DFS. 这道题的难点在于如何存储岛屿的形状. 我们定义dfs函数, 将pos列表初始化为空, 然后递归求得岛屿的其他部分相对于左上角的偏移量, 将偏移量存入pos, 将pos转化为元组放入集合中, 最后求集合的长度即可.

代码

class Solution:
    def numDistinctIslands(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        def dfs(x, y, pos, rel_pos):
            if grid[x][y] != 1:
                return
            grid[x][y] = -1
            directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]
            for dx, dy in directions:
                if 0 <= x+dx < row and 0 <= y+dy < col and grid[x+dx][y+dy] == 1:
                    new_rel_pos = (rel_pos[0] + dx, rel_pos[1] + dy)
                    pos.append(new_rel_pos)
                    dfs(x+dx, y+dy, pos, new_rel_pos)
            
        shapes = set()
        row, col = len(grid), len(grid[0])
        for x in range(row):
            for y in range(col):
                if grid[x][y] == 1:
                    # get the shape of island
                    pos = []
                    dfs(x, y, pos, (0, 0))
                    shapes.add(tuple(pos))
                    
        return len(shapes)