148.Sort List

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Linked List, Easy

Question

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

Example 1:

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

Example 2:

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

Answer

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
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);
    }
    public ListNode findMid(ListNode head){
        ListNode fast = head, slow = head, mid = null;
        while(fast!=null && fast.next!=null){
            mid = slow;
            fast = fast.next.next;
            slow = slow.next;
        }
        return mid;
    }
    public ListNode merge(ListNode l1, ListNode l2){
        ListNode dummy = new ListNode(0);
        ListNode head = dummy;
        while(l1!=null && l2!=null){
            if(l1.val < l2.val){
                head.next = l1;
                l1 = l1.next;
            }else{
                head.next = l2;
                l2 = l2.next;
            }
            head = head.next;
        }
        if(l1!=null)
            head.next = l1;
        else 
            head.next = l2;
        return dummy.next;
    }
}

Using fast and slow pointer to find middle element in LinkedList

t = length/2

slow position = t v = length/2 1 = length/2

Running time: O(nlgn)