Question

Formatted question description: https://leetcode.ca/all/147.html

147	Insertion Sort List

Sort a linked list using insertion sort.
https://upload.wikimedia.org/wikipedia/commons/0/0f/Insertion-sort-example-300px.gif

Algorithm of Insertion Sort:
    1. Insertion sort iterates, consuming one input element each repetition,
        and growing a sorted output list.
    2. 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.
    3. It repeats until no input elements remain.

Example 1:

Input: 4->2->1->3
Output: 1->2->3->4

Example 2:

Input: -1->5->3->4->0
Output: -1->0->3->4->5


@tag-linkedlist

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

Java

  • 
    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;
    	      }
    	  }
    
    }
    
  • // 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;
        }
    };
    
  • 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
    
    

All Problems

All Solutions