[劍指offer] 複雜連結清單的複制

題目描述

輸入一個複雜連結清單（每個節點中有節點值，以及兩個指針，一個指向下一個節點，另一個特殊指針指向任意一個節點），傳回結果為複制後複雜連結清單的head。（注意，輸出結果中請不要傳回參數中的節點引用，否則判題程式會直接傳回空）

解題思路

image

參考代碼

/*
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;
        //複制節點 A->B->C 變成 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;
        }
        //複制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;
    }
}