*Sort List

题目：

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

思路：

考虑到要求用O(nlogn)的时间复杂度和constant space complexity来sort list，自然而然想到了merge sort方法。同时我们还已经做过了merge k sorted list和merge 2 sorted list。这样这个问题就比较容易了。

不过这道题要找linkedlist中点，那当然就要用最经典的faster和slower方法，faster速度是slower的两倍，当faster到链尾时，slower就是中点，slower的next是下一半的开始点。

解决了这些问题，题目就很容易解出了。

代码：

public class Solution {
    public ListNode sortList(ListNode head) 
    {
        if(head==null||head.next==null) return head;
        
        ListNode fast = head;
        ListNode slow = head;
        ListNode firsthalf = head;
        
        while(fast.next!=null&&fast.next.next!=null)
        {
            slow = slow.next;
            fast = fast.next.next;
        }
        
        ListNode secondhalf = slow.next;
        slow.next=null;
        
        ListNode rightlist = null;
        ListNode leftlist = null;
        if(firsthalf!=secondhalf)
        {
            rightlist = sortList(secondhalf);
            leftlist = sortList(firsthalf);
        }
        
        return mergeList(rightlist,leftlist);
        
    }
    
    public ListNode mergeList(ListNode rightlist, ListNode leftlist)
    {
        if(rightlist==null)
        {
            return leftlist;
        }
        
        if(leftlist==null)
        {
            return rightlist;
        }
        
        ListNode fakehead = new ListNode(0);
        ListNode ptr = fakehead;
        
        while(rightlist!=null&&leftlist!=null)
        {
            if(rightlist.val<leftlist.val)
            {
                ptr.next=rightlist;
                ptr=ptr.next;
                rightlist=rightlist.next;
            }
            else
            {
                ptr.next=leftlist;
                ptr=ptr.next;
                leftlist=leftlist.next;
                
            }
        }
        
        if(rightlist!=null)ptr.next = rightlist;
        if(leftlist!=null)ptr.next = leftlist;
        return fakehead.next;

    }
       
}

 http://www.cnblogs.com/springfor/p/3869372.html