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

一刷
题解：merge sort.
注意，merge时，最后多出来的部分，直接用p.next = l1就可以。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode sortList(ListNode head) {
        if(head == null || head.next == null) return head;
        
       // step 1. cut the list to two halves
        ListNode prev = null, slow = head, fast = head;
        
        while(fast!=null && fast.next!=null){
            prev = slow;
            slow = slow.next;
            fast = fast.next.next;
        }
        
        prev.next = null;
        
        //step 2. sort each half
        ListNode l1 = sortList(head);
        ListNode l2 = sortList(slow);
        
        return merge(l1, l2);
    }
    
    private ListNode merge(ListNode l1, ListNode l2){
        ListNode l = new ListNode(0), p = l;
        
        while(l1 !=null && l2 != null){
            if(l1.val < l2.val){
                p.next = l1;
                l1 = l1.next;
            }
            else{
                p.next = l2;
                l2 = l2.next;
            }
            p = p.next;
        }
        
        if(l1!=null) p.next = l1;
        if(l2!=null) p.next = l2;
        return l.next;
    }
}

二刷
merge sort

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode sortList(ListNode head) {
        if(head == null || head.next == null) return head;
        
        //step 1: cut the list to two halves
        ListNode prev = null, slow = head, fast = head;
        while(fast!=null && fast.next!=null){
            prev = slow;
            slow = slow.next;
            fast = fast.next.next;
        }
        
        prev.next = null;
        
        //sort each half
        ListNode l1 = sortList(head);
        ListNode l2 = sortList(slow);
        
        return merge(l1, l2);
    }
    
    private ListNode merge(ListNode l1, ListNode l2){
        ListNode dummy = new ListNode(0);
        ListNode p = dummy;
        while(l1!=null && l2!=null){
            if(l1.val<l2.val){
                p.next = l1;
                l1 = l1.next;
            }
            else{
                p.next = l2;
                l2 = l2.next;
            }
            p = p.next;
        }
        if(l1!=null) p.next = l1;
        if(l2!=null) p.next = l2;
        return dummy.next;
    }
}