随笔记录
算法-链表-复制带随机指针的链表
2022-4-5 diaba
package com.jiucaiyuan.net.algrithm.linked;

import java.util.HashMap;

/**
* 【问题】复制带有随机指针节点的链表
* 【要求】时间复杂度O(N),空间复杂度O(1)
* 【思路】如果不是空间复杂度O(1)的限制,可以使用Map,暂存新老节点之间的关系,
* 然后再补齐rand字段,但是有O(1)限制,所以在克隆出新节点时,放到该节点之后,
* 变成注入这个结构:1——>1`——>2——>2`——>3——>3`——>4——>4`——>5——>5`
* 然后再遍历next、next.next,补齐rand,然后再拆出来新老链表,返回新链表的头结点
*
* @Author jiucaiyuan 2022/4/5 21:46
* @mail services@jiucaiyuan.net
*/

public class CopyListWithRand {
/** 使用额外空间
* @param head
* @return
*/
public static Node copyListWithRand1(Node head, int value) {
if (head == null){
return null;
}
//把所有原节点放到map的key中,并克隆节点作为map的value
HashMap<Node,Node> map = new HashMap<>();
Node curr = head;
while (curr != null){
map.put(curr, new Node(curr.value));
curr = curr.next;
}

//循环设置新节点的next和rand
curr = head;
while (curr != null){
//curr 老的
//map.get(curr) 新的
map.get(curr).next = map.get(curr.next);
map.get(curr).rand = map.get(curr.rand);
curr = curr.next;
}
return map.get(head);
}

/** 不使用额外空间
* @param head
* @return
*/
public static Node copyListWithRand2(Node head, int value) {
if (head == null){
return null;
}
//copy node : 1——>2——>3——>null to 1——>1`——>2——>2`——>3——>3`——>null
Node curr = head;
Node next = null;
while (curr != null){
next = curr.next;
curr.next = new Node(curr.value);
curr.next.next = next;
}

curr = head;
Node currCopy = null;
//set copy node rand
//1——>1`——>2——>2`——>3——>3`——>null
while (curr != null){
next = curr.next.next;
currCopy = curr.next;
currCopy.rand = next.rand != null ? next.rand.next:null;
curr = next;
}

Node res = head.next;
curr = head;
//split 1——>1`——>2——>2`——>3——>3`——>null to 1——>2——>3——>null and 1`——>2`——>3`——>null
while (curr != null){
next = curr.next.next;
currCopy = curr.next;
curr.next = next;
currCopy.next = next != null ? next.next:null;
curr = next;
}
return res;
}

static class Node{
int value;
Node next;
Node rand;
public Node(int value){
this.value = value;
}
}
}
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