复杂链表的复制

Posted 2020-09-17 望川拓

public class Solution {
public RandomListNode Clone(RandomListNode pHead){
if(pHead == null)
return null;
RandomListNode pCur = pHead;
//复制next 如原来是A->B->C 变成A->A\'->B->B\'->C->C\'
while(pCur!=null){
RandomListNode node = new RandomListNode(pCur.label);
node.next = pCur.next;
pCur.next = node;
pCur = node.next;
}
pCur = pHead;
//复制random pCur是原来链表的结点 pCur.next是复制pCur的结点
while(pCur != null){
if(pCur.random != null)
pCur.next.random = pCur.random.next;
pCur = pCur.next.next;
}
RandomListNode head = pHead.next;
RandomListNode cur = head;
pCur = pHead;
//拆分链表
while(pCur != null){
pCur.next = pCur.next.next;
if(cur.next != null)
cur.next = cur.next.next;
cur = cur.next;
pCur = pCur.next;
}
return head;
}
}