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

Solution (Java)

/**
 * 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 = head;
        ListNode fast = head;
        ListNode pre = slow;
        while (fast != null && fast.next != null) {
            pre = slow;
            slow = slow.next;
            fast = fast.next.next;
        }
        pre.next = null;
        ListNode first = sortList(head);
        ListNode second = sortList(slow);
        ListNode res = new ListNode(1);
        ListNode cur = res;
        while (first != null || second != null) {
            if (first == null) {
                cur.next = second;
                break;
            } else if (second == null) {
                cur.next = first;
                break;
            } else if (first.val <= second.val) {
                cur.next = first;
                first = first.next;
                cur = cur.next;
            } else {
                cur.next = second;
                second = second.next;
                cur = cur.next;
            }
        }
        return res.next;
    }
}

Solution (Javascript)

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var sortList = function(head) {
  if (!head || !head.next) return head;
  var slow = head;
  var fast = head;
  var prev = null;
  while (fast && fast.next) {
    prev = slow;
    slow = slow.next;
    fast = fast.next.next;
  }
  prev.next = null;
  return merge(sortList(head), sortList(slow));
};

var merge = function (list1, list2) {
  var p1 = list1;
  var p2 = list2;
  var newHead = new ListNode(0);
  var now = newHead;
  while (p1 || p2) {
    if (!p1 || !p2) {
      now.next = p1 || p2;
      break;
    } else if (p1.val < p2.val) {
      now.next = p1;
      p1 = p1.next;
    } else {
      now.next = p2;
      p2 = p2.next;
    }
    now = now.next;
    now.next = null;
  }
  return newHead.next;
};

Explain:

nope.

Complexity:

• Time complexity : O(nlog(n)).
• Space complexity : O(n).