leetCode---linked list topic

Topic Description 1 Singly Linked List Sorting Problem

Sort a linked list in O(n log n) time using constant space complexity.

With merge sort: where only a preHead node is created takes space O(1) time O(nlogn)

public class Solution {
    public ListNode sortList(ListNode head) {
        if(head == null || head.next == null) {
            return head;
        } //Conventional merge sort idea
        ListNode mid = getMid(head);
        ListNode midNext = mid.next;
        mid.next = null; //Be sure to disconnect the tail of the linked list in the left half of the head
        return mergeSort (sortList (head), sortList (midNext));
    }
    private ListNode getMid(ListNode head) { //Use the fast and slow pointer to get the position of the middle node
        if(head == null || head.next == null) {
            return head;
        }
        ListNode slow = head, quick = head;
        while(quick.next != null && quick.next.next != null) {
            slow = slow.next;
            quick = quick.next.next;
        }
        return slow;
    }
    private ListNode mergeSort(ListNode n1, ListNode n2) { Merge two sub-linked lists
        ListNode preHead = new ListNode(0), cur1 = n1, cur2 = n2, cur = preHead;
        while(cur1 != null && cur2 != null) {
            if(cur1.val < cur2.val) {
                cur.next = cur1;
                cur1 = cur1.next;
            } else {
                cur.next = cur2;
                cur2 = cur2.next;
            }
            cur = cur.next;
        }
        cur.next = cur1 == null ? cur2 : cur1; //See which linked list is not empty and continue to connect
        return preHead.next;
    }
}

Use quicksort:

public class Solution {
    public ListNode sortList(ListNode head) {
        quickSort(head, null); //head and tail nodes
        return head;
    }
 
    public static void quickSort(ListNode head, ListNode end) {
        if(head != end) {
            ListNode

      partition= partition(head);

        quickSort(head, partition); //分解成两部分        quickSort(partition.next, end);        }    }     public static ListNode partition(ListNode head) {        ListNode slow = head;        ListNode fast = head.next;        while (fast != null) {            if(fast.val < head.val) {                slow = slow.next;                fast.val = slow.val ^ fast.val ^ (slow.val = fast.val);            }            fast = fast.next;        }        slow.val = head.val ^ slow.val ^ (head.val = slow.val);        return slow;    }}