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;
    }
}