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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0001_0100.s0086_partition_list;

// #Medium #Two_Pointers #Linked_List #2022_06_20_Time_1_ms_(62.66%)_Space_43_MB_(25.29%)

import com_github_leetcode.ListNode;

/*
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
/**
 * 86 - Partition List\.
 *
 * Medium
 *
 * 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:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/01/04/partition.jpg)
 *
 * **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`
**/
public class Solution {
    public ListNode partition(ListNode head, int x) {
        ListNode nHead = new ListNode(0);
        ListNode nTail = new ListNode(0);
        ListNode ptr = nTail;
        ListNode temp = nHead;
        while (head != null) {
            ListNode nNext = head.next;
            if (head.val < x) {
                nHead.next = head;
                nHead = nHead.next;
            } else {
                nTail.next = head;
                nTail = nTail.next;
            }
            head = nNext;
        }
        nTail.next = null;
        nHead.next = ptr.next;
        return temp.next;
    }
}