LeetCode 1042. Flower Planting With No Adjacent 染色

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解决一个图论问题，即如何在给定的花园网络中选择不同类型的花朵种植，以确保相连的花园不会种植相同类型的花。通过创建一个图，并使用一种算法来为每个花园分配一种颜色，从而满足题目要求。

You have N gardens, labelled 1 to N. In each garden, you want to plant one of 4 types of flowers.

paths[i] = [x, y] describes the existence of a bidirectional path from garden x to garden y.

Also, there is no garden that has more than 3 paths coming into or leaving it.

Your task is to choose a flower type for each garden such that, for any two gardens connected by a path, they have different types of flowers.

Return any such a choice as an array answer, where answer[i] is the type of flower planted in the (i+1)-th garden. The flower types are denoted 1, 2, 3, or 4. It is guaranteed an answer exists.

Example 1:

Input: N = 3, paths = [[1,2],[2,3],[3,1]]
Output: [1,2,3]

Example 2:

Input: N = 4, paths = [[1,2],[3,4]]
Output: [1,2,1,2]

Example 3:

Input: N = 4, paths = [[1,2],[2,3],[3,4],[4,1],[1,3],[2,4]]
Output: [1,2,3,4]

Note:

1 <= N <= 10000
0 <= paths.size <= 20000
No garden has 4 or more paths coming into or leaving it.
It is guaranteed an answer exists.

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刚开始想把相邻节点用DFS染色成不一样，发现写起来代码有些长。换个思路，先根据已经染色的点挑一个颜色把自己染了，其他节点染色的时候会根据双向图回避自身的颜色

class Solution:
    def select_available(self, colors):
        for i in range(1,5):
            if (i not in colors):
                return i
        
    def gardenNoAdj(self, N: int, paths: List[List[int]]) -> List[int]:
        res,graph = [0 for i in range(N)],[[] for i in range(N)]
        for path in paths:
            x,y = path[0]-1,path[1]-1
            graph[x].append(y)
            graph[y].append(x)
        for i in range(N):
            res[i] = self.select_available({res[j] for j in graph[i]})
        return res