[leetcode#148] Sort List

The problem:

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

My Analysis:

The idea behind this problem is easy:merge sort!
But we should learn some tricky skills from the this question.
1. How to split a linked list into separate linked lists?
A. Use Pointers:walker and runner.
A.1 Walker move one step in each iteration.
A.2 runner move, steps in each iteration.
B. Iterate until runner can ' t move any more (Runner.next = = NULL | | runner.next.next = NULL)
C. Then head1 are the head of original list, and head2 is the next element of runner.

2. How to merge the linked list in an elegant?
The key idea behind writing a elegant code is always use the same invarint.
To achieve this purpose, I use a dummy head as fake head and a pre Pinter always point to the last element of the Resultin G list.
ListNode dummy = new ListNode (0);
Pre = dummy;
...
while () {
Pre.next = ...;
}
return dummy.next;

My Code:

/*** Definition for singly-linked list. * Class ListNode {* int val; * ListNode Next; * ListNode (int x) {* val = x; * next = NULL; *     } * } */ Public classSolution { PublicListNode sortlist (ListNode head) {returnMergeSort (head); }        PrivateListNode mergesort (ListNode head) {if(Head = =NULL|| Head.next = =NULL)//The base case in the recursion is very important!            returnHead; ListNode Walker=Head; ListNode Runner=Head;  while(Runner.next! =NULL&& Runner.next.next! =NULL) {//Skill:check Runner.next at first.Walker= Walker.next;//This skill is amazing!!!Runner =Runner.next.next; } ListNode head2=Walker.next; Walker.next=NULL; ListNode Head1=Head; Head1=mergesort (HEAD1); Head2=mergesort (head2); Head=merge (Head1, head2); returnHead; }        Privatelistnode Merge (ListNode head1, ListNode head2) {if(Head1 = =NULL&& Head2 = =NULL)            return NULL; if(Head1 = =NULL&& head2! =NULL)            returnhead2; if(Head1! =NULL&& Head2 = =NULL)            returnHead1; ListNode Dummy=NewListNode (0); ListNode Pre=dummy; ListNode ptr1=Head1; ListNode PTR2=head2;  while(Ptr1! =NULL&& PTR2! =NULL) {                        if(Ptr1.val <=ptr2.val) {Pre.next=ptr1; Pre=Pre.next; PTR1=Ptr1.next; } Else{Pre.next=ptr2; Pre=Pre.next; PTR2=Ptr2.next; }        }                if(PTR1 = =NULL) Pre.next=ptr2; ElsePre.next=ptr1; returnDummy.next; }}

[leetcode#148] Sort List