*复杂链表的复制

转大神解法：

解题思路：

1、遍历链表，复制每个结点，如复制结点A得到A1，将结点A1插到结点A后面；

2、重新遍历链表，复制老结点的随机指针给新结点，如A1.random = A.random.next;

3、拆分链表，将链表拆分为原链表和复制后的链表

public class Solution {

    public RandomListNode Clone(RandomListNode pHead) {

        if(pHead == null) {

            return null;

        }

         

        RandomListNode currentNode = pHead;

        //1、复制每个结点，如复制结点A得到A1，将结点A1插到结点A后面；

        while(currentNode != null){

            RandomListNode cloneNode = new RandomListNode(currentNode.label);

            RandomListNode nextNode = currentNode.next;

            currentNode.next = cloneNode;

            cloneNode.next = nextNode;

            currentNode = nextNode;

        }

         

        currentNode = pHead;

        //2、重新遍历链表，复制老结点的随机指针给新结点，如A1.random = A.random.next;

        while(currentNode != null) {

            currentNode.next.random = currentNode.random==null?null:currentNode.random.next;

            currentNode = currentNode.next.next;

        }

         

        //3、拆分链表，将链表拆分为原链表和复制后的链表

        currentNode = pHead;

        RandomListNode pCloneHead = pHead.next;

        while(currentNode != null) {

            RandomListNode cloneNode = currentNode.next;

            currentNode.next = cloneNode.next;

            cloneNode.next = cloneNode.next==null?null:cloneNode.next.next;

            currentNode = currentNode.next;

        }

         

        return pCloneHead;

    }

}