JZ-025-Copy of complex linked list

Replication of complex linked lists

Title description

Enter a complex linked list (each node has a node value and two pointers, one points to the next node, and the other special pointer random points to a random node), please make a deep copy of this linked list,

• And return to the head node after copying. (Attention, please do not return the node reference in the parameter in the output result, otherwise the judgment program will directly return empty)

title link : replication of complex linked list

Code

/**
 * 标题：复杂链表的复制
 * 题目描述
 * 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random指向一个随机节点），请对此链表进行深拷贝，
 * 并返回拷贝后的头结点。（注意，输出结果中请不要返回参数中的节点引用，否则判题程序会直接返回空）
 * 题目链接：
 * https://www.nowcoder.com/practice/f836b2c43afc4b35ad6adc41ec941dba?tpId=13&&tqId=11178&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking
 */
public class Jz25 {

    /**
     * 方法：
     * 第一步，在每个节点的后面插入复制的节点。
     * 第二步，对复制节点的 random 链接进行赋值。
     * 第三步，拆分。
     *
     * @param pHead
     * @return
     */
    public RandomListNode clone(RandomListNode pHead) {
        if (pHead == null) {
            return null;
        }
        // 插入新节点
        RandomListNode cur = pHead;
        while (cur != null) {
            RandomListNode clone = new RandomListNode(cur.label);
            clone.next = cur.next;
            cur.next = clone;
            cur = clone.next;
        }
        // 建立 random 链接
        cur = pHead;
        while (cur != null) {
            RandomListNode clone = cur.next;
            if (cur.random != null) {
                clone.random = cur.random.next;
            }
            cur = clone.next;
        }
        // 拆分
        cur = pHead;
        RandomListNode pCloneHead = pHead.next;
        while (cur.next != null) {
            RandomListNode next = cur.next;
            cur.next = next.next;
            cur = next;
        }
        return pCloneHead;
    }
}

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