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题目：（简单）

LeetCode的Python执行用时随缘，只要时间复杂度没有明显差异，执行用时一般都在同一个量级，仅作参考意义。

解法一（情景模拟+哈希表）：

def gardenNoAdj(self, N: int, paths: List[List[int]]) -> List[int]:    hashmap = {
   }    for path in paths:        if path[0] not in hashmap:            hashmap[path[0]] = [path[1]]        else:            hashmap[path[0]].append(path[1])        if path[1] not in hashmap:            hashmap[path[1]] = [path[0]]        else:            hashmap[path[1]].append(path[0])    color = [0 for _ in range(N + 1)]    for i in range(1, N + 1):        if i not in hashmap:            color[i] = 1        else:            apt = [1, 2, 3, 4]            for near in hashmap[i]:                if color[near] != 0 and color[near] in apt:                    apt.remove(color[near])            color[i] = apt[0]    return color[1:]

解法二（将哈希表改为数组）：

def gardenNoAdj(self, N: int, paths: List[List[int]]) -> List[int]:    hashmap = [[] for _ in range(N + 1)]    for path in paths:        if path[0] > path[1]:            hashmap[path[0]].append(path[1])        else:            hashmap[path[1]].append(path[0])    color = [0 for _ in range(N + 1)]    for i in range(1, N + 1):        apt = [1, 2, 3, 4]        for near in hashmap[i]:            if color[near] != 0 and color[near] in apt:                apt.remove(color[near])        color[i] = apt[0]    return color[1:]

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