148. Sort List

/**
 * 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) {
        ListNode result = head;
        ListNode small = head;
        ListNode path = head;
        
        while (path.next != null) {
            if (path.next.val < path.val) {
                small.next = path.next;
                path.next = path.next.next;
            }
            path = path.next;
        }
        
        ListNode fix = small;
        
        while (small.next != null) {
            
        }
        
    }
}

해보다가 현타와서.. 루션아~^^

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

    ListNode merge(ListNode list1, ListNode list2) {
        ListNode dummyHead = new ListNode();
        ListNode tail = dummyHead;
        while (list1 != null && list2 != null) {
            if (list1.val < list2.val) {
                tail.next = list1;
                list1 = list1.next;
                tail = tail.next;
            } else {
                tail.next = list2;
                list2 = list2.next;
                tail = tail.next;
            }
        }
        tail.next = (list1 != null) ? list1 : list2;
        return dummyHead.next;
    }

    ListNode getMid(ListNode head) {
        ListNode midPrev = null;
        while (head != null && head.next != null) {
            midPrev = (midPrev == null) ? head : midPrev.next;
            head = head.next.next;
        }
        ListNode mid = midPrev.next;
        midPrev.next = null;
        return mid;
    }
}

루션이 속도 실화야?