单向链表归并排序 Java

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;
        
    }
}