题链接：https://leetcode-cn.com/problems/fu-za-lian-biao-de-fu-zhi-lcof/

分三步：

（1）复制每一个节点，使得复制后的节点都在当前节点的下一个节点

（2）原生链表的节点的指向任意节点，使复制的节点也都指向某一任意节点

（3）重新连接节点，把原生节点重新连接起来，把克隆后的节点连接起来

写链表的题，尽量把链表结构和迭代过程画出来，可以清晰的看出迭代过程，也更容易找出错误。

var copyRandomList = function(head) {
    if (head == null) {
        return null;
    }
    //克隆节点
    let tempHead = head;
    while (tempHead != null) {
        let cloneNode = new Node(tempHead.val);
        let temp = tempHead.next;
        tempHead.next = cloneNode;
        cloneNode.next = temp;
        tempHead = cloneNode.next;
    }
    
    //克隆Random指针
    tempHead = head;
    while (tempHead != null) {
        let cloneNode = tempHead.next;
        if (tempHead.random != null) {
            let temp = tempHead.random;
            cloneNode.random = temp.next;
        }
        tempHead = cloneNode.next;
    }
    
    //将克隆节点连接起来
    let tempNode = head.next;
    let resNode = tempNode;
    head.next = tempNode.next;
    head = head.next;
    while (head != null) {
        tempNode.next = head.next;
        head.next = head.next.next;
        tempNode = tempNode.next;
        head = head.next;
    }

    return resNode;
};

假设链表长度为n：

时间复杂度：O(n)

空间复杂度：O(1)