148. Sort List
https://leetcode.com/problems/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---
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
/** | |
* 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) { | |
// Base case => terminate | |
if (head == null || head.next == null) { | |
return head; | |
} | |
ListNode prev = null; | |
ListNode slow = head; | |
ListNode fast = head; | |
while (fast != null && fast.next != null) { | |
prev = slow; | |
slow = slow.next; | |
fast = fast.next.next; | |
} | |
prev.next = null; | |
ListNode l1 = sortList(head); | |
ListNode l2 = sortList(slow); | |
return merge(l1, l2); | |
} | |
private ListNode merge(ListNode l1, ListNode l2) { | |
ListNode curr = new ListNode(-1); | |
ListNode ans = curr; | |
while (l1 != null && l2 != null) { | |
if (l1.val < l2.val) { | |
curr.next = l1; | |
l1 = l1.next; | |
} else { | |
curr.next = l2; | |
l2 = l2.next; | |
} | |
curr = curr.next; | |
} | |
if (l1 != null) { | |
curr.next = l1; | |
} | |
if (l2 != null) { | |
curr.next = l2; | |
} | |
return ans.next; | |
} | |
} |