Leetcode-sort List

inch using constant space complexity.

The time complexity requirement of this problem is O (n logn), it is very easy to think of the solution with MergeSort.

/*** Definition for singly-linked list. * Class ListNode {* int val; * ListNode Next; * ListNode (int x) {* val = x; * next = NULL; *     } * } */  Public classSolution { Publiclistnode Merge (ListNode head1, ListNode head2) {ListNode dummy=NewListNode (-1); ListNode Tail=dummy;  while(Head1! =NULL&& head2! =NULL){            if(Head1.val <head2.val) {Tail.next=Head1; Head1=Head1.next; }            Else{Tail.next=head2; Head2=Head2.next; } Tail=Tail.next; }        if(Head1! =NULL) {Tail.next=Head1; }        if(Head2! =NULL) {Tail.next=head2; }        returnDummy.next; }             PublicListNode Findmid (ListNode head) {ListNode slow=Head; ListNode Fast= Head.next;//instead of = head to ensure slow are in the mid position         while(Fast! =NULL&& Fast.next! =NULL) {Slow=Slow.next; Fast=Fast.next.next; }        returnslow; }         PublicListNode sortlist (ListNode head) {if(Head = =NULL|| Head.next = =NULL){            returnHead; } ListNode Mid=Findmid (head); ListNode Right=sortlist (Mid.next); Mid.next=NULL; ListNode Left=Sortlist (head); returnmerge (Left,right); }    }

It is worth noting that the

1. In the merge function, dummy the pre-node and tail,

2. and findmin ListNode fast = head.next role,

function of Mid.next = NULL in 3.sortList

Leetcode-sort List