题解 | #复杂链表的复制#

```function RandomListNode(x){
    this.label = x;
    this.next = null;
    this.random = null;
}
function Clone(pHead)
{
    // write code here
    //思路：复制插入节点，赋值random，拆分节点
    if(pHead===null){return null}
    //复制插入节点
    let temp = pHead
    while(temp){
        let copyNode =new RandomListNode(temp.label)
        copyNode.next = temp.next
        temp.next = copyNode
        temp = temp.next.next
    }
    //赋值复制节点的random
    temp = pHead
    while(temp){
        if(temp.random){
            temp.next.random = temp.random.next //复制节点紧邻原节点，复制节点的random节点也紧邻原节点的random节点
        } else{temp.next.random = null}
        temp = temp.next.next
    }
    //拆分节点
    let root = pHead.next //取出原链表偶数节点依次赋给新链表的节点
    let r = root
    temp = pHead.next
    while(temp){
        if(temp.next===null){break} 
        temp = temp.next.next//此处需要先判断是否为最后的节点，若没有判断：temp.next为null时，解析null.next报错
        r.next = temp
        r = r.next
    }
    //返回新链表头节点
    return root
}
module.exports = {
    Clone : Clone
};