给定一个链表和一个特定值 x，对链表进行分隔，使得所有小于 x 的节点都在大于或等于 x 的节点之前。

你应当保留两个分区中每个节点的初始相对位置。

 

示例:

输入: head = 1->4->3->2->5->2, x = 3
输出: 1->2->2->4->3->5

/**
 * @param {ListNode} head
 * @param {number} x
 * @return {ListNode}
 */
var partition = function (head, x) {
  const dummyHead1 = {
    next: null,
  };
  const dummyHead2 = {
    next: null,
  };

  let current = {
    next: head,
  };
  let currentL1 = dummyHead1;
  let currentL2 = dummyHead2;
  while (current.next) {
    current = current.next;
    if (current.val < x) {
      currentL1.next = current;
      currentL1 = current;
    } else {
      currentL2.next = current;
      currentL2 = current;
    }
  }

  currentL2.next = null;

  currentL1.next = dummyHead2.next;

  return dummyHead1.next;
};

class Solution:
    def partition(self, head: ListNode, x: int) -> ListNode:
        """在原链表操作，思路基本一致，只是通过指针进行区分而已"""
        # 在链表最前面设定一个初始node作为锚点，方便返回最后的结果
        first_node = ListNode(0)
        first_node.next = head
        # 设计三个指针，一个指向小于x的最后一个节点，即前后分离点
        # 一个指向当前遍历节点的前一个节点
        # 一个指向当前遍历的节点
        sep_node = first_node
        pre_node = first_node
        current_node = head

        while current_node is not None:
            if current_node.val < x:
                # 注意有可能出现前一个节点就是分离节点的情况
                if pre_node is sep_node:
                    pre_node = current_node
                    sep_node = current_node
                    current_node = current_node.next
                else:
                    # 这段次序比较烧脑
                    pre_node.next = current_node.next
                    current_node.next = sep_node.next
                    sep_node.next = current_node
                    sep_node = current_node
                    current_node = pre_node.next
            else:
                pre_node = current_node
                current_node = pre_node.next

        return first_node.next

class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
        ListNode dummy, geHead, *ltTail = &dummy, *geTail = &geHead;
        while (head) {
            auto p = head;
            head = head->next;
            if (p->val < x) {
                ltTail->next = p;
                ltTail = p;
            } else {
                geTail->next = p;
                geTail = p;
            }
        }
        ltTail->next = geHead.next;
        geTail->next = NULL;
        return dummy.next;
    }
};