M1: 复制带随机指针的链表

一个链表问题：复制带随机指针的链表

题目:有一个链表L,其每个节点有2个指针，一个指针next指向链表的下个节点，另一个random随机指向链表中的任一个节点，可能是自己或者为空，写一个程序，要求复制这个链表的结构并分析其复杂性

解决方法一：

O(n)的复杂度,扫面两边即可。

图【1】
图【1】是需要复制的链表

图【2】

如图【2】所示，ABCD是原来的链表，A’B’C’D’是复制的链表，第一遍扫描顺序复制next指针，把ABCD的next分别指向A’B’C’D’，将A’的next指针指向B，B’的next指针指向C，依次类推

复制random指针： A’->random=A->random->next

恢复:A->next=A’->next;A’->next=A’->next->next;

package careercup;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;

public class RandomList {
	public static RandomList single = new RandomList();

	public RandomList() {
	}

	public static void main(String[] args) {
		int[] arr = new int[]{1,3,5,7};
		RandomNode head = RandomNode.createList(arr); 
		head.randomNext = (RandomNode)head.next.next;
		((RandomNode)head.next).randomNext = (RandomNode)head.next.next.next;
		
		RandomNode rn = single.getList(head);
		
		printList(rn);
		//printList(head);
	}

	private static void printList(RandomNode rn) {
		while(rn!=null){
			System.out.print(rn);
			System.out.print(", ");
			rn = (RandomNode)rn.next;
		}
		
	}

	public RandomNode getList(RandomNode rn) {
		RandomNode head = rn;
		RandomNode p=null;
		
		while(rn!=null){
			p = new RandomNode(rn);
			p.next = rn.next;
			rn.next = p;
			rn = (RandomNode)p.next;
		}
		
		p = (RandomNode)head.next;
		while(p!=null){
			p.randomNext = p.randomNext!=null ? (RandomNode)p.randomNext.next : null;
			if(p.next!=null) p = (RandomNode)p.next.next;
			else break;
		}
		
		rn = head;
		head = (RandomNode)head.next;
		p = head;
		rn.next = p.next;
		while(rn.next!=null){
			p.next = (RandomNode)p.next.next;
			p = (RandomNode)p.next;
			rn.next = p.next;
		}
		return head;
	}
}

class RandomNode extends Node{
	RandomNode randomNext;
	
	RandomNode(int val) {
		super(val);
		randomNext = null;
	}
	
	RandomNode(RandomNode rn){
		super(rn.val);
		this.next = rn.next;
		this.randomNext = rn.randomNext;
	}
	
	public static RandomNode createList(int[] arr) {
		RandomNode head = new RandomNode(arr[0]);
		RandomNode p = head;
		for(int i=1; i<arr.length;i++){
			p.next = new RandomNode(arr[i]);
			p = (RandomNode)p.next;
		}	
		return head;
	}
	
	public String toString(){
		return "[" + val + "(" +  randomNext + ")]"; 
	}
}