Insertion Sort List
Problem Statement :
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
Solution :
Solution in C :
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* insertionSortList(struct ListNode* head){
struct ListNode* new_head=NULL;
struct ListNode* ptr;
struct ListNode* temp;
if (head==NULL) return head;
new_head = head;
head = head->next;
new_head->next = NULL;
while (head!=NULL) {
temp = head;
head = head->next;
ptr = new_head;
while ((ptr->next!=NULL) && (ptr->next->val < temp->val))
ptr = ptr->next;
if (ptr->val > temp->val) {
temp->next = new_head;
new_head = temp;
} else {
temp->next = ptr->next;
ptr->next = temp;
}
}
return new_head;
}
Solution in C++ :
/**
* 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(-1, head);
ListNode curr = head;
ListNode numInsert, prev;
while(curr != null && curr.next != null) {
if(curr.val <= curr.next.val) {
curr = curr.next;
} else {
numInsert = curr.next;
prev = dummy;
while(prev.next.val < numInsert.val) {
prev = prev.next;
}
curr.next = numInsert.next;
numInsert.next = prev.next;
prev.next = numInsert;
}
}
return dummy.next;
}
}
Solution in Java :
class Solution {
public ListNode insertionSortList(ListNode head) {
if(head == null || head.next == null){
return head;
}
ListNode mid = getMid(head);
ListNode left = insertionSortList(head);
ListNode right = insertionSortList(mid);
return merge(left,right);
}
ListNode merge(ListNode list1, ListNode list2) {
ListNode dummyHead = new ListNode();
ListNode tail = dummyHead;
while (list1 != null && list2 != null) {
if (list1.val < list2.val) {
tail.next = list1;
list1 = list1.next;
tail = tail.next;
} else {
tail.next = list2;
list2 = list2.next;
tail = tail.next;
}
}
tail.next = (list1 != null) ? list1 : list2;
return dummyHead.next;
}
ListNode getMid(ListNode head) {
ListNode midPrev = null;
while (head != null && head.next != null) {
midPrev = (midPrev == null) ? head : midPrev.next;
head = head.next.next;
}
ListNode mid = midPrev.next;
midPrev.next = null;
return mid;
}
}
Solution in Python :
class Solution:
def insertionSortList(self, head: Optional[ListNode]) -> Optional[ListNode]:
temp=head
res=[]
while(temp):
res.append(temp.val)
temp=temp.next
res.sort()
print(res)
new=head
i=0
while new: # fill the sorted values
new.val = res[i]
i += 1
new = new.next
return head
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