【leetcode】1042. Flower Planting With No Adjacent

题目如下：

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.

解题思路：可供选的花的种类只有[1,2,3,4]四种，对于任意一个待种植的花园，只需要判断相邻的花园是否已经种植花卉。如果种植了，把已种植的种类从可供选择的列表中去除，最后在剩余的种类中任选一个即可。

代码如下：

class Solution(object):
    def gardenNoAdj(self, N, paths):
        """
        :type N: int
        :type paths: List[List[int]]
        :rtype: List[int]
        """
        res = [0] * (N+1)
        res[1] = 1
        dic = {}
        for v1,v2 in paths:
            dic[v1] = dic.setdefault(v1,[]) + [v2]
            dic[v2] = dic.setdefault(v2,[]) + [v1]
        for i in range(2,N+1):
            if i not in dic:
                res[i] = 1
            else:
                choice = [1,2,3,4]
                for neibour in dic[i]:
                    if res[neibour] == 0:
                        continue
                    else:
                        if res[neibour] in choice:
                            inx =  choice.index(res[neibour])
                            del choice[inx]
                res[i] = choice[0]
        return res[1:]