148. Sort List

Problem:

Sort a linked list in O(n log n) time using constant space complexity.

Example 1:

Input: 4->2->1->3
Output: 1->2->3->4

Example 2:

Input: -1->5->3->4->0
Output: -1->0->3->4->5

Analysis:
Merge sort实现题。最底层递归返回条件是ListNode 只剩下一个了，就返回它自己。
findMid还是用fast 从head.next开始。

class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null)
            return head;
        ListNode mid = findMid(head);
        ListNode right = sortList(mid.next);
        mid.next = null;
        ListNode left = sortList(head);
        return merge(left, right);
    }

    private ListNode findMid(ListNode head) {
        ListNode slow = head;
        ListNode fast = head.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }

    private ListNode merge(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode p = dummy;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                p.next = l1;
                l1 = l1.next;
            } else {
                p.next = l2;
                l2 = l2.next;
            }
            p = p.next;
         }

        if (l1 != null) {
            p.next = l1;
        } else {
            p.next = l2;
        }
        return dummy.next;
    }
}