题解 | #单链表的排序#

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 *   public ListNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

/**
 * NC70 单链表的排序
 * @author d3y1
 */
public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param head ListNode类 the head node
     * @return ListNode类
     */
    public ListNode sortInList (ListNode head) {
        // return solution1(head);
        return solution2(head);
        // return solution3(head);
    }

    /**
     * 堆排序
     * @param head
     * @return
     */
    private ListNode solution1(ListNode head){
        // PriorityQueue<ListNode> minHeap = new PriorityQueue<>((o1, o2) -> (o1.val-o2.val));
        PriorityQueue<ListNode> minHeap = new PriorityQueue<>(Comparator.comparingInt(o -> o.val));

        while(head != null){
            minHeap.offer(new ListNode(head.val));
            head = head.next;
        }

        ListNode top;
        ListNode tail = new ListNode(-1);
        if(!minHeap.isEmpty()){
            top = minHeap.poll();
            head = top;
            tail = top;
        }

        while(!minHeap.isEmpty()){
            top = minHeap.poll();
            tail.next = top;
            tail = tail.next;
        }

        return head;
    }

    /**
     * 归并排序(自顶向下)
     * @param head
     * @return
     */
    private ListNode solution2(ListNode head){
        return mergeSort(head);
    }

    /**
     * 归并
     * @param head
     * @return
     */
    private ListNode mergeSort(ListNode head){
        if(head==null || head.next==null){
            return head;
        }

        ListNode dummyHead = new ListNode(-1);
        dummyHead.next = head;
        ListNode fast = dummyHead;
        ListNode slow = dummyHead;
        while(fast!=null && fast.next!=null){
            fast = fast.next.next;
            slow = slow.next;
        }

        ListNode left = head;
        ListNode right = slow.next;
        slow.next = null;

        return merge(mergeSort(left), mergeSort(right));
    }

    /**
     * 合并
     * @param left
     * @param right
     * @return
     */
    private ListNode merge(ListNode left, ListNode right){
        ListNode tail = new ListNode(-1);
        ListNode head = tail;

        while(left!=null && right!=null){
            if(left.val <= right.val){
                tail.next = left;
                left = left.next;
            }else{
                tail.next = right;
                right = right.next;
            }
            tail = tail.next;
        }

        tail.next = (left!=null) ? left : right;

        return head.next;
    }

    /**
     * 归并排序(自底向上)
     * @param head
     * @return
     */
    public ListNode solution3(ListNode head) {
        if(head == null){
            return head;
        }
        
        int len = 0;
        ListNode node = head;
        while(node != null){
            len++;
            node = node.next;
        }
        
        ListNode dummyHead = new ListNode(-1);
        dummyHead.next = head;
        for(int subLen=1; subLen<len; subLen<<=1){
            ListNode prev = dummyHead;
            ListNode curr = dummyHead.next;
            while(curr != null){
                ListNode left = curr;
                for(int i=1; i<subLen&&curr.next!=null; i++){
                    curr = curr.next;
                }
                
                ListNode right = curr.next;
                curr.next = null;
                curr = right;
                for(int i=1; i<subLen&&curr!=null&&curr.next!=null; i++){
                    curr = curr.next;
                }
                
                ListNode next = null;
                if(curr != null){
                    next = curr.next;
                    curr.next = null;
                }
                
                ListNode merged = merge(left, right);
                prev.next = merged;
                while(prev.next != null){
                    prev = prev.next;
                }
                curr = next;
            }
        }
        
        return dummyHead.next;
    }
}