Welcome to Subscribe On Youtube
Question
Formatted question description: https://leetcode.ca/all/147.html
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
Algorithm
One by one element is taken out of the original linked list and then inserted into the new linked list in order.
The time complexity is O(n^2), which is an algorithm that is not very efficient, but the space complexity is O(1), in exchange for high time complexity for low space complexity.
Code
-
public class Insertion_Sort_List { /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode insertionSortList(ListNode head) { if (head == null) { return null; } // for current tail, maintain its prev and itself, for moving around ListNode dummy = new ListNode(0); // dummy.next = head; // @note:@memorize: here is the start of wrong /* @note: I think the key is to cut the list into 2 half, un-linked I was trying to move node around using i, j to indicate the progress of sorted part, like insert-sort in arrays but it was nightmare moving pointers the key is the same as the below solution which create new nodes but here no need to create new nodes, just pointer moving and key is the same, keep 2 separate, un-linked, un-related lists */ ListNode current = head; while (current != null) { // every time, scanner will scan from begging of sorted list, to find where to insert ListNode scanner = dummy; // @note:@memorize: here use scanner.next, so that insert postion is at scanner next while (scanner.next != null && scanner.next.val < current.val) { scanner = scanner.next; } // record for next loop ListNode nextRoundStart = current.next; // insert current.next = scanner.next; scanner.next = current; // update current = nextRoundStart; } return dummy.next; } } // create a separate list, extra space... public class Solution_extra_space { public ListNode insertionSortList(ListNode head) { if (head == null || head.next == null) return head; ListNode dummy = new ListNode(0); dummy.next = new ListNode(head.val); ListNode current = head.next; // add head before loop while(current != null) { ListNode prev = dummy; ListNode newnode = new ListNode(current.val); while(prev.next != null && newnode.val > prev.next.val) { prev = prev.next; } // insert ListNode oldPrevNext = prev.next; prev.next = newnode; newnode.next = oldPrevNext; current = current.next; // loop through the list } 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; } } -
// OJ: https://leetcode.com/problems/insertion-sort-list/ // Time: O(N^2) // Space: O(1) class Solution { public: ListNode* insertionSortList(ListNode* head) { ListNode dummy; while (head) { auto node = head; head = head->next; auto p = &dummy; while (p->next && p->next->val < node->val) p = p->next; node->next = p->next; p->next = node; } return dummy.next; } }; -
# 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 ############ class Solution(object): def insertionSortList(self, head): p = dummy = ListNode(0) cur = dummy.next = head while cur and cur.next: val = cur.next.val if cur.val < val: cur = cur.next continue if p.next.val > val: p = dummy while p.next.val < val: p = p.next new = cur.next cur.next = new.next new.next = p.next p.next = new 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; };