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【LeetCode with Python】 138. Copy List with Random Pointer

题目

原题页面:https://leetcode.com/problems/copy-list-with-random-pointer/
本文地址:</copy-list-with-random-pointer/>
题目类型:Hash Table, Linked List
难度评价:Hard
类似题目:(M) Clone Graph

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.


分析


代码

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# Definition for singly-linked list with a random pointer.
class RandomListNode:
def __init__(self, x):
self.label = x
self.next = None
self.random = None

class Solution:
# @param head, a RandomListNode
# @return a RandomListNode
def copyRandomList(self, head):
if None == head:
return None
save_list = [ ]
p1 = head
while None != p1:
save_list.append(p1)
p1 = p1.next
new_head = RandomListNode(-1)
new_head.next = head
first = new_head
second = head
copy_head = RandomListNode(-1)
copy_first = copy_head
copy_second = None
while None != first:
copy_second = RandomListNode(second.label) if None != second else None
copy_first.next = copy_second
copy_first = copy_first.next
first = first.next
if None != second:
second = second.next

p1 = head
p1_tail = head.next
p2 = copy_head.next
while None != p1:
p1_tail = p1.next
p1.next = p2
p2.random = p1
p1 = p1_tail
p2 = p2.next
p2 = copy_head.next
#p1 = head
while None != p2:
p2.random = p2.random.random.next if None != p2.random.random else None
#p1.next = p1.next.next.random if None != p1.next.next else None # may broken the previous p1.next, so have to save ori list
p2 = p2.next
#p1 = p1.next
len_save_list = len(save_list)
for i in range(0, len_save_list - 1):
save_list[i].next = save_list[i + 1]
save_list[len_save_list - 1].next = None
return copy_head.next