直接上答案了,挺没意思的这道。
总的来说就是,先在每一个点后面直接插相同的,然后处理细节,最后再分开提出来刚才复制的
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;
}
}
i小宋