Sort List

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

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public  ListNode findMid(ListNode head) {
        ListNode pSlow = head;
        ListNode pFast = head;
        while (pFast.next!=null && pFast.next.next !=null) {
            pSlow = pSlow.next;
            pFast = pFast.next.next;
        }
        return pSlow;
    }

    public  ListNode sortList(ListNode head) {
        if (head==null||head.next==null){
            return head;
        }
        ListNode pMid = findMid(head);
        ListNode pMidRight = pMid.next;
        pMid.next = null;
        return merge(sortList(head),sortList(pMidRight));
    }

    public  ListNode merge(ListNode list1, ListNode list2) {
        if (list1==null || list2==null){
            return list1==null ?list2 : list1;
        }
        ListNode result;
        ListNode p1 = list1;
        ListNode p2 = list2;
        //初始化头部
        if (p1.val < p2.val) {
            result = p1;
            p1 = p1.next;
        } else {
            result = p2;
            p2 = p2.next;
        }
        ListNode p = result;
        //合并
        while (p1!=null && p2!=null) {
            if (p2.val < p1.val) {
                p.next =  p2;
                p2 = p2.next;
                p = p.next;
            } else {
                p.next = p1;
                p1 = p1.next;
                p = p.next;
            }
        }
        //补齐
        while (p1!=null) {
            p.next = p1;
            p1 = p1.next;
            p = p.next;
        }
        while (p2!=null) {
            p.next = p2;
            p2 = p2.next;
            p = p.next;
        }
        return result;
    }

}