Sort List

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

思路：merge sort Linked List, jame bone,找到middle point的前一个，然后把list 劈开；

/**
 * 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;
        } 
        ListNode slow = findMiddle(head);
        ListNode newhead = slow.next;
        slow.next = null;
        
        ListNode l1 = sortList(head);
        ListNode l2 = sortList(newhead);
        return merge(l1, l2);
    }
    
    private ListNode findMiddle(ListNode head) {
        ListNode dummpy = new ListNode(-1);
        dummpy.next = head;
        
        ListNode slow = dummpy;
        ListNode fast = dummpy;
        
        while(fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
    
    private ListNode merge(ListNode l1, ListNode l2) {
        ListNode dummpy = new ListNode(-1);
        ListNode cur = dummpy;
        while(l1 != null && l2 != null) {
            if(l1.val < l2.val) {
                cur.next = l1;
                l1 = l1.next;
                cur = cur.next;
            } else {
                cur.next = l2;
                l2 = l2.next;
                cur = cur.next;
            }
        }
        if(l1 != null) {
            cur.next = l1;
        }
        if(l2 != null) {
            cur.next = l2;
        }
        return dummpy.next;
    }
}