LeetCode -- Sort List

Question:

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

Analysis:

问题描述：在O(n log n)时间内用常数的空间复杂度，完成一个链表的排序。

思考：排序时间复杂度为O(n log n)的话，快排、堆排序、归并排序都是O(n log n)，而冒泡排序、交换排序、选择排序都是O(n2)，就这道题来讲，很容易相当应该使用归并排序。

之前做过一个题目合并两个已经排好序的链表，因此这里的思路是：递归+归并排序。每次将当前链表分成两部分，排序后合并成一个链表。（注意归并的结束条件）

Answer：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
  public ListNode sortList(ListNode head) {
        if(head == null || head.next == null)
                return head;
        ListNode slow = head, fast = head, firstList = head;
        
        while(fast.next != null && fast.next.next != null) {
                slow = slow.next;
                fast = fast.next.next;
        }
         ListNode secondList = slow.next;
         slow.next = null;
         
         ListNode first = null, second = null;
         while(firstList != secondList) {
                 first = sortList(firstList);
                 second = sortList(secondList);
                 return mergeTwoLists(first, second);
         }
         
        return mergeTwoLists(first, second);
    }
    
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode p1 = l1;
        ListNode p2 = l2;
        
        if(l1 == null && l2 == null)
            return null;
        if(l1 == null && l2 != null)
            return l2;
        if(l1 != null && l2 == null)
            return l1;
        
        ListNode fakeHead = new ListNode(0);
        ListNode p = fakeHead;
        
        while (p1 != null && p2 != null){
            if (p1.val <= p2.val) {
                p.next = p1;
                p1 = p1.next;
                p = p.next;
            } else {
                p.next = p2;
                p2 = p2.next;
                p = p.next;
            }
        }
        
        if(p1 != null)
            p.next = p1;
        if(p2 != null)
            p.next = p2;
        return fakeHead.next;
    }

}