题目来源:LeetCode(力扣)
给定一个链表,每个节点包含一个额外增加的随机指针,该指针可以指向链表中的任何节点或空节点。
要求返回这个链表的深拷贝。
思路:
分三步走,首先,将新老链表相互间隔的串为一个链表;然后,处理random指针域;最后,将老新链表拆开,并返回对应的链表头结点。
/*
// Definition for a Node.
class Node {
public int val;
public Node next;
public Node random;
public Node() {}
public Node(int _val,Node _next,Node _random) {
val = _val;
next = _next;
random = _random;
}
};
*/
class Solution {
public Node copyRandomList(Node head) {
if(head == null){
return null;
}
//1、将新老结点串为一个链表
Node cur = head;
while(cur != null){
Node node = new Node(cur.val,cur.next,null);
Node tmp = cur.next;
cur.next = node;
cur = tmp;
}
//2、处理random指针域
cur = head;
while(cur != null){
if(cur.random != null){
cur.next.random = cur.random.next;
cur = cur.next.next;
}else{
cur.next.random = null;
cur = cur.next.next;
}
}
//3、将老新链表拆开,返回新链表
cur = head;
Node newHead = cur.next;
while(cur.next != null){
Node tmp = cur.next;
cur.next = tmp.next;
cur = tmp;
}
return newHead;
}
}