148. 排序链表

考察归并排序，快慢指针得中间节点 先切割，分治，然后合并排序 O(n log n) 时间复杂度，常数级空间复杂度

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null) 
            return head;
        // split the list into two halfs
        ListNode left = head;
        ListNode right = getMid(head);
        ListNode tmp = right.next;
        right.next = null;
        right = tmp;

        // 分治
        left = sortList(left);
        right = sortList(right); 

        // 合并
        return merge(left, right);
    }

    // 用快慢指针算法来得到链表的中点
    private ListNode getMid(ListNode head) {
        ListNode slow = head, fast = head.next;
        while(fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }

    // 合并
    private ListNode merge(ListNode list1 , ListNode list2) {
        ListNode tail = new ListNode(), dummy = tail;
        while(list1 != null && list2 != null) {
            if(list1.val < list2.val) {
                tail.next = list1;
                list1 = list1.next;
            }
            else {
                tail.next = list2;
                list2 = list2.next;
            }
            tail = tail.next;
        }
        if (list1 != null)
            tail.next = list1; // 已经有序，直接尾插整个list1即可

        if (list2 != null) 
            tail.next = list2;

        return dummy.next;
    }
}