LeetCode-剑指 Offer 35. 复杂链表的复制

思路：使用haspmap存储原来的结点和新节点的对应

代码：

"""
# Definition for a Node.
class Node:
    def __init__(self, x: int, next: 'Node' = None, random: 'Node' = None):
        self.val = int(x)
        self.next = next
        self.random = random
"""
class Solution:
    def copyRandomList(self, head: 'Node') -> 'Node':
        if not head: return None
        hashmap, p = {}, head
        while p:
            hashmap[p] = Node(p.val)
            p = p.next
        p = head
        while p:
            hashmap[p].next = hashmap.get(p.next)
            hashmap[p].random = hashmap.get(p.random)
            p = p.next
        return hashmap[head]