[leetcode]138. Copy List with Random Pointer@Java解题报告

https://leetcode.com/problems/copy-list-with-random-pointer/description/

A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.

Return a deep copy of the list.

package go.jacob.day804;import java.util.HashMap;import java.util.Map;/** * 138. Copy List with Random Pointer 题意是：返回一个链表的深拷贝 *  * @author Jacob * */public class Demo2 {/* * 推荐解答：使用map,每一个原链表节点，在map中对应新链表的节点 */public RandomListNode copyRandomList(RandomListNode head) {if (head == null)return null;Map<RandomListNode, RandomListNode> map = new HashMap<RandomListNode, RandomListNode>();RandomListNode node = head;while (node != null) {map.put(node, new RandomListNode(node.label));node = node.next;}node = head;while (node != null) {map.get(node).next = map.get(node.next);map.get(node).random = map.get(node.random);node = node.next;}return map.get(head);}/* * 剑指offer中的解答，会改变原链表的结构。 需要额外操作来恢复原链表的结构 * 链表的题目画图比较清晰 */public RandomListNode copyRandomList_1(RandomListNode head) {if (head == null)return null;RandomListNode node = head;// 在原链表每个节点后插入一个复制的节点while (node != null) {RandomListNode newNode = new RandomListNode(node.label);RandomListNode temp = node.next;node.next = newNode;newNode.next = temp;node = temp;}node = head;// 复制随机节点while (node != null) {RandomListNode tempRandom = node.random;if (tempRandom != null) {node.next.random = tempRandom.next;} elsenode.next.random = null;node = node.next.next;}//恢复原链表和生成新链表RandomListNode newHead = head.next;RandomListNode newNode = head.next;node = head;while (newNode != null) {node.next=newNode.next;node = newNode.next;if (node != null){newNode.next = node.next;}elsenewNode.next = null;newNode = newNode.next;}return newHead;}private class RandomListNode {int label;RandomListNode next, random;RandomListNode(int x) {this.label = x;}};}