排序链表

题目来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/sort-list

在 O(n log n) 时间复杂度和常数级空间复杂度下，对链表进行排序。

示例 1:

输入: 4->2->1->3
输出: 1->2->3->4

示例 2:

输入: -1->5->3->4->0
输出: -1->0->3->4->5

归并解法：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode sortList(ListNode head) {
        return sortAndMerge(head);
    }

    /**
     * 归并排序
     * @param head
     * @return
     */
    private ListNode sortAndMerge(ListNode head){
        if (head == null)
            return null;
        ListNode fast = head.next;
        ListNode slow = head;
        while(fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode rightNode = slow.next;
        slow.next = null;
        //此时slow节点就是中间节点，按这个节点分割
        ListNode left = head;
        if (head.next != null){
            left = sortAndMerge(head);
        }
        ListNode right = sortList(rightNode);
        return merge(left,right);
    }

    /**
     * 合并两个有序链表
     *  1->3
     *  2->4
     *  1->2->3->4
     * @param l1
     * @param l2
     * @return
     */
    private ListNode merge(ListNode l1,ListNode l2){
        ListNode head = new ListNode(0);
        ListNode cur = new ListNode(0);
        head.next = cur;
        while(l1 != null || l2 != null){
            if (l1 == null){
                cur.next = l2;
                break;
            }
            if (l2 == null){
                cur.next = l1;
                break;
            }
            if (l1.val > l2.val){
                cur.next = l2;
                l2 = l2.next;
            }else{
                cur.next = l1;
                l1 = l1.next;
            }
            cur = cur.next;
        }
        return head.next.next;
    }
}