边工作边刷题：70天一遍leetcode: day 11

Copy List with Random Pointer

思路很容易理解，这题连接完了cur.next是永远存在的，所以copy random link的时候不用特殊处理。break因为一轮包括4个结点，另外在list结尾有特殊情况，所以比较复杂

• 顺序：以当前（cur）和下一个（被copy结点）为中心，去连接这两个的next，然后只需要沿cur.next推进就可以
• 特殊情况：因为被copy结点的next是null，不能连接其下下个结点。所以最后一个copy.next是no-op即可

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

class Solution(object):
    def copyRandomList(self, head):
        """
        :type head: RandomListNode
        :rtype: RandomListNode
        """
        if not head: return None
        cur = head
        while cur:
            next = cur.next
            cur.next = RandomListNode(cur.label)
            cur.next.next = next
            cur = next
        
        cur = head
        while cur:
            if cur.random:
                cur.next.random = cur.random.next
            cur = cur.next.next
        
        cur = head
        newH = head.next
        while cur:
            copy = cur.next
            cur.next = copy.next
            if cur.next:
                copy.next = cur.next.next
            cur=cur.next
        return newH