Leetcode: Sort List

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

 记得Insert Sort List, 那个复杂度是O(N^2)的，这里要求O（nlogn），所以想到merge sort, 需要用到Merge Two Sorted List的方法（我写的merge函数）

第二遍代码：所以mergesort不好写在于它是一个递归里嵌套另一个递归，第一个递归不停地用runner technique把list分两段，直到每一段是一个或0个节点返回该点，这时再调用merge Two sorted List递归把两段整合起来，返回它们的首节点

public class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null) return head;
        return mergesort(head);
    }
    
    public ListNode mergesort(ListNode head) {
        if (head == null || head.next == null) return head;
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode walker = dummy;
        ListNode runner = dummy;
        while (runner!=null && runner.next!=null) {
            runner = runner.next.next;
            walker = walker.next;
        }
        ListNode head1 = dummy.next;
        ListNode head2 = walker.next;
        walker.next = null;
        return merge(mergesort(head1), mergesort(head2));
    }
    
    public ListNode merge(ListNode head1, ListNode head2) {
        if (head1 == null) return head2;
        if (head2 == null) return head1;
        ListNode dummy = new ListNode(-1);
        dummy.next = head1;
        ListNode pre = dummy;
        while (head1 != null && head2 != null) {
            if (head1.val <= head2.val) {
                head1 = head1.next;
            }
            else {
                ListNode next = head2.next;
                head2.next = pre.next;
                pre.next = head2;
                head2 = next;
            }
            pre = pre.next;
        }
        if (head2 != null) {
            pre.next = head2;
        }
        return dummy.next;
    }
}