leetcode1042 Flower Planting With No Adjacent
作者:互联网
1 """ 2 You have N gardens, labelled 1 to N. In each garden, you want to plant one of 4 types of flowers. 3 paths[i] = [x, y] describes the existence of a bidirectional path from garden x to garden y. 4 Also, there is no garden that has more than 3 paths coming into or leaving it. 5 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. 6 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. 7 Example 1: 8 Input: N = 3, paths = [[1,2],[2,3],[3,1]] 9 Output: [1,2,3] 10 Example 2: 11 Input: N = 4, paths = [[1,2],[3,4]] 12 Output: [1,2,1,2] 13 Example 3: 14 Input: N = 4, paths = [[1,2],[2,3],[3,4],[4,1],[1,3],[2,4]] 15 Output: [1,2,3,4] 16 """ 17 """ 18 此题用贪心算法 19 没掌握 20 """ 21 class Solution(object): 22 def gardenNoAdj(self, N, paths): 23 res = [0] * N 24 m = [[] for i in range(N + 1)] # 用来存第i个结点到其他结点的路径,i为1到N 25 for x, y in paths: 26 m[x].append(y) 27 m[y].append(x) 28 for i in range(1, N + 1): # 对N个路径进行遍历 29 used = set() # 用来存这个结点出现的所有路径使用过的颜色 30 for j in m[i]: 31 used.add(res[j - 1]) # 把目的地的颜色放入used里 32 for j in range(1, 5): # 1,2,3,4代表四种颜色 33 if j not in used: 34 res[i - 1] = j # 将该结点存入新的颜色 35 break 36 return res
标签:Planting,结点,Flower,garden,paths,res,leetcode1042,used,Example 来源: https://www.cnblogs.com/yawenw/p/12257560.html