题目描述
输入一个复杂链表(每个节点中有节点值,以及两个指针,一个指向下一个节点,另一个特殊指针指向任意一个节点),返回结果为复制后复杂链表的head。(注意,输出结果中请不要返回参数中的节点引用,否则判题程序会直接返回空)
思路
两种方案
- hashMap快速复制,空间复杂度O(N)
- 在原链表上复制结点,并拆分
代码实现
import java.util.HashMap;
public class Offer35_Clone {
// 定义复杂链表
public static class RandomNode{
int val;
RandomNode next;
RandomNode random;
RandomNode(int data){
this.val = data;
}
}
// 1.hash解决方案
public RandomNode clone(RandomNode head){
if(head == null) return null;
// 构建一个hashMap,key为Node,value为Node的复制值
HashMap<RandomNode, RandomNode> hash = new HashMap<>();
RandomNode cur = head;
while(cur != null){
hash.put(cur, new RandomNode(cur.val));
cur = cur.next;
}
// 给每个复制的结点赋值next和random指针
cur = head;
while(cur != null){
hash.get(cur).next = hash.get(cur.next);
hash.get(cur).random = hash.get(cur.random);
cur = cur.next;
}
return hash.get(head);
}
public static RandomNode copy2(RandomNode head){
if(head == null) return null;
// 在原链表的每个结点后面复制一个结点并连接成新链表
RandomNode cur = head;
RandomNode next = null;
while(cur != null){
next = cur.next;
cur.next = new RandomNode(cur.val);
cur.next.next = next;
cur = next;
}
// 对复制结点设置random指针,指向当前复制结点前一个结点的random指针的下一个节点
cur = head;
RandomNode curCopy = null;
while(cur != null){
next = cur.next.next;
curCopy = cur.next;
curCopy.random = cur.random != null ? cur.random.next: null;
cur = next;
}
// 分离
RandomNode res = head.next;
cur = head;
while(cur != null){
next = cur.next.next;
curCopy = cur.next;
curCopy.next = next != null ? next.next: null;
cur = next;
}
return res;
}
}