## 138. Copy List with Random Pointer

Published on with 0 views and 0 comments

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`.

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.

golang：

``````/**
* Definition for a Node.
* type Node struct {
*     Val int
*     Next *Node
*     Random *Node
* }
*/

}

// build newList
var mapping = make(map[*Node]*Node, 0)
for temp:= head; temp != nil; temp = temp.Next{
mapping[temp] = &Node{Val: temp.Val}

}

// 拷贝next指针和random指针关系
for temp:= head; temp != nil; temp = temp.Next {
mapping[temp].Next = mapping[temp.Next]
mapping[temp].Random = mapping[temp.Random]
}

}

// O(1) 空间的迭代
}

// 1. 把复制的节点都放在被复制的节点后
for temp:= head; temp != nil; {
newNode := &Node{Val:temp.Val, Next: temp.Next}
temp.Next = newNode
temp = newNode.Next
}

// 2. 复制random指针关系
for temp:= head; temp != nil; temp = temp.Next.Next {
if temp.Random != nil {
// 因为 next节点保存的都是新节点，随机指针指向的下一个节点肯定也是新链表的下一个节点
temp.Next.Random = temp.Random.Next
}
}
// 3. 和拆分奇偶链表一样拆开单链表就行
var (
)
for even != nil && even.Next != nil {
odd.Next = even.Next
odd = odd.Next // 实际上是even

even.Next = odd.Next
even = even.Next
}
odd.Next = nil // 抹掉odd链上最后一个节点的next

return newList
}