Partition List leetcode

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.

这道题我本想用交换节点的方法处理，因为势必要保存当前节点的前一个节点，所以交换起来非常复杂

后来上网一查，发现可以直接使用创建两个链表然后合并的方法，不禁恍然大悟，忽略了链表的灵活性

既然partition方法可以很容易的实现，那么链表的快速排序也可以实现了，有时间再实现

ListNode* partition(ListNode* head, int x) {
    if (head == nullptr)
        return head;
    ListNode temp1(0);
    ListNode temp2(0);
    ListNode *cur1 = &temp1;
    ListNode *cur2 = &temp2;
    temp1.next = head;
    int tag = 0;
    while (head != nullptr)
    {
        if (head->val < x)
        {            
            if (tag == 1) 
            {
                cur1->next = head;
                tag = 0;
            }
            cur1 = cur1->next;
        }
        else if (head->val >= x)
        {        
            if (tag == 0)
            {
                cur2->next = head;
                tag = 1;
            }
            cur2 = cur2->next;        
        }
        head = head->next;
    }
    cur1->next = cur2->next = nullptr;
    cur1->next = temp2.next;
    return temp1.next;
}