## [86\. Partition List](https://leetcode.com/problems/partition-list/)

Given the `head` of 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.

**Example 1:**


**Input:** head = \[1,4,3,2,5,2\], x = 3
**Output:** \[1,2,2,4,3,5\]

**Example 2:**

**Input:** head = \[2,1\], x = 2
**Output:** \[1,2\]

**Constraints:**

- The number of nodes in the list is in the range `[0, 200]`.
- `-100 <= Node.val <= 100`
- `-200 <= x <= 200`

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
        // 建立兩條 linked list: node1, node2;
        ListNode* node1 = new ListNode(-1);
        ListNode* node2 = new ListNode(-1);
        ListNode* ptr1 = node1;
        ListNode* ptr2 = node2;
        while (head) {
            // 如果 head 的 val 比 x 小
            if (head->val < x) {
                // 將 head 接到 ptr1->next
                ptr1->next = head;
                // 移動 ptr1
                ptr1 = ptr1->next;
            } else {
                // 反之，接到 ptr2
                ptr2->next = head;
                // 移動 ptr2
                ptr2 = ptr2->next;
            }
            // 移動 head
            head = head->next;
        }
        // ptr2->next 為尾端 NULL
        ptr2->next = nullptr;
        // ptr1 後面接 node2->next
        ptr1->next = node2->next;
        // 最後返回 node1->next
        return node1->next;
    }
};

:::success
- 時間複雜度：$O()$
- 空間複雜度：$O()$
:::