Description
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:
- Insertion sort iterates, consuming one input element each repetition and growing a sorted output list.
- At each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted list and inserts it there.
- It repeats until no input elements remain.
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:
- The number of nodes in the list is in the range
[1, 5000].
-5000 <= Node.val <= 5000
Solutions
Solution 1
PythonJavaJavaScriptGo
# 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(code-box)
/**
* 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;
}
}(code-box)
/**
* 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;
};(code-box)
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func insertionSortList(head *ListNode) *ListNode {
if head == nil || head.Next == nil {
return head
}
dummy := &ListNode{head.Val, head}
pre, cur := dummy, head
for cur != nil {
if pre.Val <= cur.Val {
pre = cur
cur = cur.Next
continue
}
p := dummy
for 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
}(code-box)
