148-排序链表Sort-List

148. 排序链表
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

分析：和这道题差不多阿，都是链表，用归并排序做。题目要求 O(nlogn)，好像就是快排和归并? 递归的终止条件是一对有序对。

class Solution {public ListNode sortList(ListNode head) {if (head == null || head.next == null)
            return head;
        ListNode fast = head.next, slow = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode tmp = slow.next;
        slow.next = null;
        ListNode left = sortList(head);
        ListNode right = sortList(tmp);
        ListNode h = new ListNode(0);
        ListNode res = h;
        while (left != null && right != null) {if (left.val < right.val) {
                h.next = left;
                left = left.next;
            } else {
                h.next = right;
                right = right.next;
            }
            h = h.next;
        }
        h.next = left != null ? left : right;
        return res.next;
    }
}