LintCode 531 [Six Degrees]

原题

六度分离是一个哲学问题，说的是每个人每个东西可以通过六步或者更少的步数建立联系。
现在给你一个友谊关系，查询两个人可以通过几步相连，如果不相连返回 -1

样例
给出图

1------2-----4
 \          /
  \        /
   \--3--/

{1,2,3#2,1,4#3,1,4#4,2,3} s = 1, t = 4 返回 2

给出图二

1      2-----4
             /
           /
          3

{1#2,4#3,4#4,2,3} s = 1, t = 4 返回 -1

解题思路

• 本题相当于求图中两点的最短路径，BFS -> 用Queue来实现
• 加入hash map去重，并且巧妙的将step当作值存入每一个节点为key的hash map中，这样通过节点A找到的相邻节点的步数即map[A] + 1

完整代码

# Definition for Undirected graph node
# class UndirectedGraphNode:
#     def __init__(self, x):
#         self.label = x
#         self.neighbors = []
import Queue

class Solution:
    '''
    @param {UndirectedGraphNode[]} graph a list of Undirected graph node
    @param {UndirectedGraphNode} s, t two Undirected graph nodes
    @return {int} an integer
    '''
    def sixDegrees(self, graph, s, t):
        # Write your code here
        members = {}
        q = Queue.Queue(maxsize = len(graph))

        q.put(s)
        members[s] = 0
        while not q.empty():
            x = q.get()
            if x == t:
                return members[x]

            for y in x.neighbors:
                if y not in members:
                    members[y] = members[x] + 1
                    q.put(y)

        return -1