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Given the head
of a singly linked list, sort the list using insertion sort, and return the sorted list's head.
The steps of the insertion sort algorithm:
The following is a graphical example of the insertion sort algorithm. The partially sorted list (black) initially contains only the first element in the list. One element (red) is removed from the input data and inserted in-place into the sorted list with each iteration.
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]
Constraints:
[1, 5000]
.-5000 <= Node.val <= 5000
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def insertionSortList(self, head: ListNode) -> ListNode:
if head is None or head.next is None:
return head
dummy = ListNode(head.val, head)
pre, cur = dummy, head
while cur:
if pre.val <= cur.val:
pre, cur = cur, cur.next
continue
p = dummy
while p.next.val <= cur.val:
p = p.next
t = cur.next
cur.next = p.next
p.next = cur
pre.next = t
cur = t
return dummy.next
/**
* 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 insertionSortList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode dummy = new ListNode(head.val, head);
ListNode pre = dummy, cur = head;
while (cur != null) {
if (pre.val <= cur.val) {
pre = cur;
cur = cur.next;
continue;
}
ListNode p = dummy;
while (p.next.val <= cur.val) {
p = p.next;
}
ListNode t = cur.next;
cur.next = p.next;
p.next = cur;
pre.next = t;
cur = t;
}
return dummy.next;
}
}
/**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} head
* @return {ListNode}
*/
var insertionSortList = function(head) {
if (head == null || head.next == null) return head;
let dummy = new ListNode(head.val, head);
let prev = dummy, cur = head;
while (cur != null) {
if (prev.val <= cur.val) {
prev = cur;
cur = cur.next;
continue;
}
let p = dummy;
while (p.next.val <= cur.val) {
p = p.next;
}
let t = cur.next;
cur.next = p.next;
p.next = cur;
prev.next = t;
cur = t;
}
return dummy.next;
};
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