138 Copy List with Random Pointer

Leetcode

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]]

---

剑指 offer 有类似的题。由于快速定位到新链表的随机节点比较困难，因此复制的时候，第一次遍历，在原链表的每个节点后面复制一个节点，第二遍遍历的时候为每个复制节点找到 random 节点，random 节点在老节点的 random 节点的下一个节点。然后再拆分链表。

/*
// Definition for a Node.
class Node {
    int val;
    Node next;
    Node random;

    public Node(int val) {
        this.val = val;
        this.next = null;
        this.random = null;
    }
}
*/

class Solution {
    public Node copyRandomList(Node head) {
        if(head==null) return null;

        Node t = head, t2;
        while(t!=null){
            Node copy=new Node(t.val);
            copy.next=t.next;
            t.next=copy;
            t=t.next.next;
        }
        t=head;
        while(t!=null){
            if(t.random!=null) t.next.random=t.random.next;
            t=t.next.next;
        }
        t=head;
        Node ret = t.next;
        t2=ret;
        
        while(true){
            t.next=t2.next;
            t=t.next;
            if(t==null) break;
            t2.next=t.next;
            t2=t2.next;
        }
        return ret;
    }
}