《剑指Offer》35. 复杂链表的复制

题目链接

牛客网

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。

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

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

解题思路

public class Solution {
    public RandomListNode Clone(RandomListNode pHead){
        if (pHead==null) return null;
        // 复制
        RandomListNode cur = pHead;
        while (cur != null) {
            RandomListNode clone = new RandomListNode(cur.label);
            clone.next = cur.next;
            cur.next = clone;
            cur = clone.next;
        }
        // 复制Random
        cur = pHead;
        while (cur != null) {
            RandomListNode clone = cur.next;
            if (cur.random != null) clone.random = cur.random.next;
            cur = clone.next;
        }
        // 将复制与原生拆开
        cur = pHead;
        RandomListNode clonePHead = pHead.next;
        while (cur.next != null) {
            RandomListNode clone = cur.next;
            cur.next = clone.next;
            cur = clone;
        }
        return clonePHead;
    }
}