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
Solution:
/**
* 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, fast = head;
ListNode prev = null;
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 head1, ListNode head2) {
if (head1 == null) return head2;
if (head2 == null) return head1;
ListNode dummy = new ListNode(0);
ListNode p1 = head1, p2 = head2;
ListNode p = dummy;
while (p1 != null && p2 != null) {
if (p1.val < p2.val) {
p.next = p1;
p1 = p1.next;
} else {
p.next = p2;
p2 = p2.next;
}
p = p.next;
}
if (p1 != null) {
p.next = p1;
}
if (p2 != null) {
p.next = p2;
}
return dummy.next;
}
}