复杂链表的复制（Java实现）

1.题目

2.解法

/*
public class RandomListNode {
    int label;
    RandomListNode next = null;
    RandomListNode random = null;

    RandomListNode(int label) {
        this.label = label;
    }
}
*/

public class Solution {
    public RandomListNode Clone(RandomListNode pHead) {
        if(pHead == null) {
            return null;
        }
        RandomListNode curNode = pHead;
        RandomListNode nextNode = null;
        //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 (curNode != null) {
            RandomListNode cloneNode = new RandomListNode(curNode.label);
            nextNode = curNode.next;
            cloneNode.next = nextNode;
            curNode.next = cloneNode;
            curNode = nextNode;
        }
       
        
        curNode = 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(curNode != null) {
            // 为什么是random.next，因为原结点后面紧跟复制的结点
            curNode.next.random = (curNode.random == null ? null : curNode.random.next);
            curNode = curNode.next.next;
        }

        //3、拆分链表，将链表拆分为原链表和复制后的链表
        curNode = pHead;
        RandomListNode cloneHead = pHead.next;
        RandomListNode cloneNode = null;
        while (curNode != null) {
            cloneNode = curNode.next;
            curNode.next = cloneNode.next;
            // 如果cloneNode.next为空的话
            cloneNode.next =  (cloneNode.next == null ? null : cloneNode.next.next);
            curNode = curNode.next;
        }
        return cloneHead;
    }
}

时间复杂度为O(3n),空间复杂度为O(n)