Sort List

很简单一道题，就是细节容易出错，必然是二分才有nlogn的速度

public class Solution {

    public ListNode sortList(ListNode head) {

        // 跟merge sort 一模一样 http://www.cnblogs.com/jiajiaxingxing/p/4397354.html

        if(head==null ||head.next==null) return head;

        ListNode h = new ListNode(-1);

        h.next = head;

        ListNode run=head, walk = head;

        while(run.next!=null && run.next.next!=null){

            run = run.next.next;

            walk=walk.next;

        }

        ListNode lft = head,rt = walk.next;

        walk.next= null;

        if(lft!=rt){ // 忘了加这个！！！！

                    lft = sortList(lft);

                    rt = sortList(rt);

        }

        

        return merge2list(lft,rt);

    }

    public ListNode merge2list(ListNode lft, ListNode rt){

        if(lft==null) return rt; // 忘了加这个！！！！

        if(rt==null) return lft; // 忘了加这个！！！！

        ListNode h = new ListNode(-1);

        ListNode cur = h;

        while(lft!=null && rt!=null){

            if(lft.val<rt.val){

                cur.next = lft;

                lft = lft.next;

                cur = cur.next;

            }else{

                cur.next = rt;

                rt = rt.next;

                cur = cur.next;

            }

        }

        if(lft!=null && rt==null) cur.next = lft;

        if(lft==null && rt!=null) cur.next = rt;

        

        return h.next;

    }

}