如何修复这个链表算法答案

【问题标题】：How to fix this algorithm of linked list如何修复这个链表算法
【发布时间】：2019-12-18 22:59:29
【问题描述】：

我想在链表上的特定元素（第二个 Joey）之后添加一个元素（steve）。

当前列表：Ash->joey->Alex->Cook->Joey->bing.

期望的输出：Ash->joey->Alex->Cook->Joey->steve->bing

这是我的代码：

def insert_at_same(self , newNode, data_to_check):
    currentNode = self.head
    temp = 0

    while currentNode.next is not None:
        if (temp == 1 and currentNode.data == data_to_check):
            tempNode  = currentNode.next
            currentNode.next = newNode
            newNode.next = tempNode
            return

        elif currentNode.data == data_to_check:
            temp = temp + 1

        else:
            currentNode = currentNode.next

但我的输出仍然是：Ash->joey->steve->Alex->Cook->Joey->bing。

【问题讨论】：

标签： python python-3.x linked-list

【解决方案1】：

通过新节点的下一个插入，而不是通过当前节点的下一个并返回头部。在这里，我在特定节点之后粘贴到插入节点.. [在 Python 中更改]

在之后/下一个插入

    SLL temp = head;
    SLL current = null;
    while(temp.next != null) {
        current = temp;
        /* if the node found, after where you need to insert */
        if(current.name.equals(nodeName)) {
            SLL newName = new SLL(name);
            /* newnode will point to second of current node */
            newName.next = current.next;
            /* initialize the new_node, to current node */
            current.next = newName;
            return head;
        }
        temp = temp.next;
    }
    /* if the node found at the last position of list */
    if(temp.name.equals(nodeName)) {
        temp.next = new SLL(name);
    }

    return head;

在之前/之前插入

    /* if the node found at first of list */
   if(head != null && head.name.equals(nodeName)) {
        SLL newName = new SLL(name);
        newName.next = head;
        head = newName;
        return head;
    }
    SLL temp = head;
    SLL current = null;
    /* traverse the list, until the node is found */
    while(temp.next != null && !temp.name.equals(nodeName)) {
        current = temp;         
        temp = temp.next;
    }

    SLL newName = new SLL(name);
    newName.next = current.next;
    current.next = newName;
    return head;

【讨论】：