Given the head of a linked list, return the list after sorting it in ascending order.
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].
- -105 <= Node.val <= 105
Follow up: Can you sort the linked list in O(n logn) time and O(1) memory (i.e. constant space)?
/**
* 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; }
* }
*/
import java.util.Collections;
class Solution {
public ListNode sortList(ListNode head) {
ListNode sorted = null;
ListNode sRef = null;
List<ListNode> list = new ArrayList<>();
while(head!=null){
list.add(new ListNode(head.val));
head = head.next;
}
Collections.sort(list, new Comparator<ListNode>(){
@Override
public int compare(ListNode o1, ListNode o2){
return o1.val-o2.val;
}
});
for(ListNode node:list){
if(sorted==null){
sorted = node;
sRef = sorted;
}else{
sRef.next = node;
sRef = sRef.next;
}
}
return sorted;
}
}二分法+两个数组合并
public class Solution {
public ListNode sortList(ListNode head) {
if (head == null || head.next == null)
return head;
// step 1. cut the list to two halves
ListNode prev = null, slow = head, fast = head;
while (fast != null && fast.next != null) {
prev = slow;
slow = slow.next;
fast = fast.next.next;
}
prev.next = null;
// step 2. sort each half
ListNode l1 = sortList(head);
ListNode l2 = sortList(slow);
// step 3. merge l1 and l2
return merge(l1, l2);
}
ListNode merge(ListNode l1, ListNode l2) {
ListNode l = new ListNode(0), 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;
}
}