68 lines
2.2 KiB
Java
68 lines
2.2 KiB
Java
# 35. 复杂链表的复制
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[NowCoder](https://www.nowcoder.com/practice/f836b2c43afc4b35ad6adc41ec941dba?tpId=13&tqId=11178&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
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## 题目描述
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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。
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```java
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public class RandomListNode {
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int label;
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RandomListNode next = null;
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RandomListNode random = null;
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RandomListNode(int label) {
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this.label = label;
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}
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}
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```
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<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/66a01953-5303-43b1-8646-0c77b825e980.png" width="300"/> </div><br>
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## 解题思路
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第一步，在每个节点的后面插入复制的节点。
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<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/dfd5d3f8-673c-486b-8ecf-d2082107b67b.png" width="600"/> </div><br>
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第二步，对复制节点的 random 链接进行赋值。
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<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/cafbfeb8-7dfe-4c0a-a3c9-750eeb824068.png" width="600"/> </div><br>
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第三步，拆分。
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<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/e151b5df-5390-4365-b66e-b130cd253c12.png" width="600"/> </div><br>
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```java
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public RandomListNode Clone(RandomListNode pHead) {
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if (pHead == null)
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return null;
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// 插入新节点
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RandomListNode cur = pHead;
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while (cur != null) {
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RandomListNode clone = new RandomListNode(cur.label);
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clone.next = cur.next;
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cur.next = clone;
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cur = clone.next;
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}
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// 建立 random 链接
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cur = pHead;
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while (cur != null) {
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RandomListNode clone = cur.next;
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if (cur.random != null)
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clone.random = cur.random.next;
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cur = clone.next;
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}
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// 拆分
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cur = pHead;
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RandomListNode pCloneHead = pHead.next;
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while (cur.next != null) {
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RandomListNode next = cur.next;
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cur.next = next.next;
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cur = next;
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}
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return pCloneHead;
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}
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```
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