单向链表归并排序 use Java
链表的关键在于递归的时候中间位置的确定，方法是：用两个指针p，f 遍历链表，p走一步而f走两步；当f走完的时候p走到链表的一半！
这让我烧绳子那道逻辑题。
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 {
public ListNode sortList(ListNode head) {
if (head == null || head.next == null){
return head;
}
ListNode p = head;
ListNode f = head.next;
while ( f.next !=null && f.next.next !=null ){//locate p at half of the ListNode
p = p.next;
f = f.next.next;
}
ListNode h2 = sortList(p.next);
p.next = null;
return merge( sortList (head) , h2 );
}
public ListNode merge(ListNode h1,ListNode h2){
ListNode hn = new ListNode(-1);
ListNode c = hn;
while (h1 != null && h2 != null ){
if (h1.val <= h2.val){
c.next = h1;
h1 = h1.next;
}else {
c.next = h2;
h2 = h2.next;
}
c = c.next;
}
if(h1 == null){
c.next = h2;
}else{
c.next = h1;
}
return hn.next;
}
}