LeetCode 148 Sort List

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

思路：要想时间复杂度达到O(n log n)
，那么有两种，一种是合并排序，另一种是快速排序，而要想空间复杂度为常数，那么只能使用递归，本人使用的是递归的合并排序，代码如下：

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
   private ListNode merge(ListNode l1, ListNode l2) {
        if(l1==null&&l2==null) return null;
        ListNode head=new ListNode (-1);
        ListNode pNode=head;
        while(l1!=null||l2!=null){
			if(l1==null) {
				pNode.next=l2;
				return head.next;
			}
			if(l2==null){
				pNode.next=l1;
				return head.next;
			}
            if(l1.val>l2.val){
				pNode.next=l2;
				l2=l2.next;
            }else {
				pNode.next=l1;
				l1=l1.next;
            }
            pNode=pNode.next;                   
        }
        return head.next;
    }
   private ListNode mergeSort(ListNode head){
       if(head==null||head.next==null)   //just one element
           return head;
       ListNode p=head, q=head, pre=null;
       while(q!=null&&q.next!=null){
           q=q.next.next;
           pre=p;
           p=p.next;  //divide into two parts
       }
       pre.next=null;
       ListNode lhalf=mergeSort(head);
       ListNode rhalf=mergeSort(p);  //recursive
       return merge(lhalf, rhalf);   //merge
   }
    public ListNode sortList(ListNode head) {
        return mergeSort( head);
    }
}