LeetCode 138 Copy List with Random Pointer

LeetCode 138 Copy List with Random Pointer

A linked list of length n is given such that each node contains an additional random pointer, which could point to any node in the list, or null.

Construct a deep copy of the list. The deep copy should consist of exactly n brand new nodes, where each new node has its value set to the value of its corresponding original node. Both the next and random pointer of the new nodes should point to new nodes in the copied list such that the pointers in the original list and copied list represent the same list state. None of the pointers in the new list should point to nodes in the original list.

For example, if there are two nodes X and Y in the original list, where X.random --> Y, then for the corresponding two nodes x and y in the copied list, x.random --> y.

Return the head of the copied linked list.

The linked list is represented in the input/output as a list of n nodes. Each node is represented as a pair of [val, random_index] where:

• val: an integer representing Node.val
• random_index: the index of the node (range from 0 to n-1) that the random pointer points to, or null if it does not point to any node.

Your code will only be given the head of the original linked list.

Example 1:

**Input:** head = [[7,null],[13,0],[11,4],[10,2],[1,0]]
**Output:** [[7,null],[13,0],[11,4],[10,2],[1,0]]

Example 2:

**Input:** head = [[1,1],[2,1]]
**Output:** [[1,1],[2,1]]

Example 3:

**Input:** head = [[3,null],[3,0],[3,null]]
**Output:** [[3,null],[3,0],[3,null]]

Example 4:

**Input:** head = []
**Output:** []
**Explanation:** The given linked list is empty (null pointer), so return null.

Constraints:

• 0 <= n <= 1000
• -10000 <= Node.val <= 10000
• Node.random is null or is pointing to some node in the linked list.

题目大意：

复制含next和random的链表。

解题思路（推荐）：

此法较容易实现。先复制next指针，然后利用HashMap存储旧新节点，来复制random指针。

注意事项：

1. 复制next指针和Map中。clone题均用此法。
2. Random指针不空才copy
3. 加it = it.next，否则死循环
4. 如果创建新Node用while it.next表示用它的父节点，否则某个field赋值如random用while it

Python代码：

def copyRandomList(self, head: 'Node') -> 'Node':
	node_map = {}
	fake_head, fake_head_copy = Node(0), Node(0)
	fake_head.next = head
	it, it_copy = fake_head, fake_head_copy
	while it.next:
		it_copy.next = Node(it.next.val)
		node_map[it.next] = it_copy.next
		it, it_copy = it.next, it_copy.next

	it, it_copy = fake_head.next, fake_head_copy.next
	while it:
		if it.random:
			node_map[it].random = node_map[it.random]
		it, it_copy = it.next, it_copy.next
	return fake_head_copy.next

梅花间竹解题思路：

第二种方法，梅花间竹，分3部走。

def copyRandomList(self, head: 'Node') -> 'Node':
	fake_head, fake_head_copy = Node(0), Node(0)
	fake_head.next = head

	# insert
	it = fake_head.next
	while it:
		temp = it.next
		it.next = Node(it.val)
		it.next.next = temp
		it = it.next.next

	# copy random
	it = fake_head.next
	while it:
		if it.random is not None:
			it.next.random = it.random.next
		it = it.next.next

	# delete
	it, it_copy = fake_head.next, fake_head_copy
	while it:
		temp = it.next
		it.next = it.next.next
		it_copy.next = temp
		temp.next = None
		it, it_copy = it.next, it_copy.next
	return fake_head_copy.next

算法分析：

1. 时间复杂度为O(n)，空间复杂度O(n)。
2. 时间复杂度为O(n)，空间复杂度O(1)。