[剑指offer] 复杂链表的复制

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。(注意,输出结果中请不要返回参数中的节点引用,否则判题程序会直接返回空)

解题思路

image

参考代码

/*
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;
        //复制节点 A->B->C 变成 A->A'->B->B'->C->C'
        RandomListNode head = pHead;
        while(head != null){
            RandomListNode node = new RandomListNode(head.label);
            node.next = head.next;
            head.next = node;
            head = node.next;
        }
        //复制random
        head = pHead;
        while(head != null){
            head.next.random = head.random == null ? null : head.random.next;
            head = head.next.next;
        }
        //折分
        head = pHead;
        RandomListNode chead = head.next;
        while(head != null){
            RandomListNode node = head.next;
            head.next = node.next;
            node.next = node.next == null ? null : node.next.next;
            head = head.next;
        }
        return chead;
    }
}

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