leetcode刷题（77）—— 148. 排序链表

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode sortList(ListNode head) {
           PriorityQueue<Integer> heap =
                new PriorityQueue<Integer>(new Comparator<Integer>() {
                    @Override
                    public int compare(Integer n1, Integer n2) {
                        return n1 - n2;
                    }
                });

        while (head != null) {
            heap.add(head.val);
            head = head.next;
        }
        ListNode pre = new ListNode(0);
        ListNode begin = new ListNode(0);
        pre = begin;
        while (!heap.isEmpty()) {
            ListNode node = new ListNode(heap.poll());
            begin.next = node;
            begin = node;
        }
        return pre.next;
    }
}

 public static ListNode sortList(ListNode head) {
        // 1、递归结束条件
        if (head == null || head.next == null) {
            return head;
        }

        // 2、找到链表中间节点并断开链表 & 递归下探
        ListNode midNode = middleNode(head);
        ListNode rightHead = midNode.next;
        midNode.next = null;

        ListNode left = sortList(head);
        ListNode right = sortList(rightHead);

        // 3、当前层业务操作（合并有序链表）
        return mergeTwoLists(left, right);
    }

    //  找到链表中间节点（876. 链表的中间结点）
    private static ListNode middleNode(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode slow = head;
        ListNode fast = head.next.next;

        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }

        return slow;
    }

    // 合并两个有序链表（21. 合并两个有序链表）
    private static ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode sentry = new ListNode(-1);
        ListNode curr = sentry;

        while(l1 != null && l2 != null) {
            if(l1.val < l2.val) {
                curr.next = l1;
                l1 = l1.next;
            } else {
                curr.next = l2;
                l2 = l2.next;
            }

            curr = curr.next;
        }

        curr.next = l1 != null ? l1 : l2;
        return sentry.next;
    }