leetcode-hard-ListNode-Copy List with Random Pointer-NO

mycode

报错：Node with val 1 was not copied but a reference to the original one.

其实我并没有弄懂对于ListNode而言咋样才叫做深拷贝了

"""
# Definition for a Node.
class Node(object):
    def __init__(self, val, next, random):
        self.val = val
        self.next = next
        self.random = random
"""
class Solution(object):
    def copyRandomList(self, head):
        """
        :type head: Node
        :rtype: Node
        """
        if not head : return None
        dummy = p = ListNode(-1)
        while head:
            p.next = head
            p = p.next
            head = head.next
        return dummy.next

 

这道链表的深度拷贝题的难点就在于如何处理随机指针的问题，由于每一个节点都有一个随机指针，这个指针可以为空，也可以指向链表的任意一个节点，如果我们在每生成一个新节点给其随机指针赋值时，都要去遍历原链表的话，OJ 上肯定会超时，所以我们可以考虑用 HashMap 来缩短查找时间，第一遍遍历生成所有新节点时同时建立一个原节点和新节点的 HashMap，第二遍给随机指针赋值时，查找时间是常数级。

 

参考

https://www.cnblogs.com/zuoyuan/p/3745126.html

1、先不考虑random的问题，在原链表上构建copy

2、在copy链表上，利用now.next.random 也需要等于now.random.next实现随即指针的拷贝

3、两个链表分离

"""
# Definition for a Node.
class Node(object):
    def __init__(self, val, next, random):
        self.val = val
        self.next = next
        self.random = random
"""
class Solution(object):
    def copyRandomList(self, head):
        """
        :type head: Node
        :rtype: Node
        """
        if head == None: return None
        tmp = head
        while tmp:
            newNode = Node(tmp.val,None,None)
            newNode.next = tmp.next
            tmp.next = newNode
            tmp = tmp.next.next
        tmp = head
        while tmp:
            if tmp.random:
                tmp.next.random = tmp.random.next
            tmp = tmp.next.next
        newhead = head.next
        pold = head
        pnew = newhead
        while pnew.next:
            pold.next = pnew.next
            pold = pold.next
            pnew.next = pold.next
            pnew = pnew.next
        pold.next = None
        pnew.next = None
        return newhead