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SortALinkedList.java
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SortALinkedList.java
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package Leetcode;
/**
* @author kalpak
*
* Given the head of a linked list, return the list after sorting it in ascending order.
*
* Follow up: Can you sort the linked list in O(n logn) time and O(1) memory (i.e. constant space)?
*
* Example 1:
* Input: head = [4,2,1,3]
* Output: [1,2,3,4]
*
*
* Example 2:
* Input: head = [-1,5,3,4,0]
* Output: [-1,0,3,4,5]
*
*
* Example 3:
* Input: head = []
* Output: []
*
* Constraints:
*
* The number of nodes in the list is in the range [0, 5 * 104].
* -10^5 <= Node.val <= 10^5
*/
public class SortALinkedList {
public static ListNode sortList(ListNode head) {
if (head == null || head.next == null)
return head;
// step 1. cut the list to two halves
ListNode prev = null;
ListNode slow = head;
ListNode fast = head;
while (fast != null && fast.next != null) {
prev = slow;
slow = slow.next;
fast = fast.next.next;
}
// Cut the lists
prev.next = null;
ListNode head2 = slow;
// step 2. sort each half
ListNode l1 = sortList(head);
ListNode l2 = sortList(head2);
// step 3. merge l1 and l2
return merge(l1, l2);
}
private static ListNode merge(ListNode l1, ListNode l2) {
ListNode l = new ListNode(0);
ListNode p = l;
while (l1 != null && l2 != null) {
if (l1.val < l2.val) {
p.next = l1;
l1 = l1.next;
}
else {
p.next = l2;
l2 = l2.next;
}
p = p.next;
}
if (l1 != null)
p.next = l1;
if (l2 != null)
p.next = l2;
return l.next;
}
public static void main(String[] args) {
ListNode l1 = new ListNode(-1);
l1.next = new ListNode(5);
l1.next.next = new ListNode(3);
l1.next.next.next = new ListNode(4);
l1.next.next.next.next = new ListNode(0);
ListNode result = sortList(l1);
while(result != null) {
System.out.print(result.val + "");
result = result.next;
}
System.out.println();
}
}