題目
原題如下
138. Copy List with Random Pointer
A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.
Return a deep copy of the list.
解釋
題目是讓大家複製一個複雜鏈表。這個鏈表爲什麼複雜呢，因爲每一個節點除了一個next指針，還多了一個隨機指針。這個隨機指針可能爲空，也可能指向任意一個節點。
解法
這個題目前沒有發現任何可以取巧的解法。必須得一步一步踏踏實實的解。先按照next指針構造好新的鏈表。在構造的同時記錄新的節點和舊的節點的對應關係，然後在依次初始化random指針。
可見思路非常簡單，但是按照這個思路寫的代碼表現如下：
Runtime: 36 ms, faster than 98.49% of Python3 online submissions for Copy List with Random Pointer.
Memory Usage: 14.7 MB, less than 100.00% of Python3 online submissions for Copy List with Random Pointer.
由此我們知道這道題彷彿沒有什麼特別取巧的解法或者思路了。
代碼如下：
"""
# Definition for a Node.
class Node:
def __init__(self, val, next, random):
self.val = val
self.next = next
self.random = random
"""
class Solution:
def copyRandomList(self, head: 'Node') -> 'Node':
if not head:
return None
headN = Node(head.val, None, None)
O2N = {head: headN}
nnPre = headN
no = head.next
while no != None:
nn = Node(no.val, None, None)
nnPre.next, nnPre = nn, nn
O2N[no] = nn
no = no.next
nn = headN
no = head
while nn:
if no.random:
nn.random = O2N[no.random]
else:
nn.random = None
nn, no = nn.next, no.next
return headN