147. Insertion Sort List.md

147. Insertion Sort List

---

leetcode Daily Challenge on November 2nd, 2020.

leetcode-cn Daily Challenge on November 20th, 2020.

---

Difficulty : Medium

Related Topics : LinkedList、Sort

---

Sort a linked list using insertion sort.

A graphical example of insertion sort. The partial sorted list (black) initially contains only the first element in the list. With each iteration one element (red) is removed from the input data and inserted in-place into the sorted list

Algorithm of Insertion Sort:

• 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.

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

---

Solution

• mine
  • Java

    • Insert Sort Runtime: 25 ms, faster than 67.04%, Memory Usage: 38.7 MB, less than 8.12% of Java online submissions

      // O(N^2)time
      // O(1)space
      public ListNode insertionSortList(ListNode head) {
          if (head == null || head.next == null) return head;
      
          ListNode t = head.next;
          head.next = null;
          while (t != null) {
              ListNode node = t;
              t = t.next;
              node.next = null;
      
              ListNode temp = head;
              if (temp.val > node.val) {
                  node.next = head;
                  head = node;
                  continue;
              }
              if (temp.next == null) {
                  temp.next = node;
                  continue;
              }
      
              ListNode pre = temp;
              ListNode cur = temp.next;
              while (cur != null) {
                  if (cur.val < node.val) {
                      if (cur.next == null) {
                          cur.next = node;
                          break;
                      } else {
                          cur = cur.next;
                          pre = pre.next;
                      }
                  } else {
                      pre.next = node;
                      node.next = cur;
                      break;
                  }
              }
          }
          return head;
      }
      

    • Runtime: 6 ms, faster than 70.63%,Memory Usage: 39.6 MB, less than 8.12% of Java online submissions

      // O(N*logN)time
      // O(N)space
      public ListNode insertionSortList(ListNode head) {
          if (head == null || head.next == null) return head;
          Map<Integer, List<ListNode>> map = new HashMap<>();
          List<Integer> list = new ArrayList<>();
          ListNode t = head;
          while (t != null) {
              ListNode temp = t;
              t = t.next;
              temp.next = null;
              List<ListNode> v = map.getOrDefault(temp.val, new ArrayList<>());
              if(v.isEmpty()) list.add(temp.val);
              v.add(temp);
              map.put(temp.val, v);
          }
          Collections.sort(list);
          ListNode res = new ListNode();
          t = res;
          for (int i = 0; i < list.size(); i++) {
              List<ListNode> temp = map.get(list.get(i));
              for (int j = 0; j < temp.size(); j++) {
                  t.next = temp.get(j);
                  t = t.next;
              }
          }
          return res.next;
      }
      

---

• the most votes

• Runtime: 31 ms, faster than 19.19%, Memory Usage: 38.3 MB, less than 8.12% of Java online submissions

  // O(N^2)time
  // O(1)space
  public ListNode insertionSortList(ListNode head) {
      if (head == null) {
          return head;
      }
  
      ListNode helper = new ListNode(0); //new starter of the sorted list
      ListNode cur = head; //the node will be inserted
      ListNode pre = helper; //insert node between pre and pre.next
      ListNode next = null; //the next node will be inserted
      //not the end of input list
      while (cur != null) {
          next = cur.next;
          //find the right place to insert
          while (pre.next != null && pre.next.val < cur.val) {
              pre = pre.next;
          }
          //insert between pre and pre.next
          cur.next = pre.next;
          pre.next = cur;
          pre = helper;
          cur = next;
      }
  
      return helper.next;
  }
  

---

• the fastest code

• MergeSort Runtime: 1 ms, faster than 99.53%, Memory Usage: 38.6 MB, less than 8.12% of Java online submissions

  public ListNode insertionSortList(ListNode head) {
      // NOTE : This is mergesort. Not insertion sort
      if(head == null || head.next == null) {
          return head;
      }
  
      ListNode slow = head;
      ListNode fast = head;
      ListNode preslow = null;
  
      while(fast != null && fast.next != null) {
          preslow = slow;
          slow = slow.next;
          fast = fast.next.next;
      }
  
      preslow.next = null;
  
      ListNode head1 = insertionSortList(head);
      ListNode head2 = insertionSortList(slow);
  
      ListNode prehead = new ListNode();
      ListNode current = prehead;
  
      while(head1!=null && head2 != null) {
          if(head1.val < head2.val){
              current.next = head1;
              head1 = head1.next;
          }else{
              current.next = head2;
              head2 = head2.next;
          }
          current = current.next;
      }
  
      if(head1!=null){
          current.next = head1;
      }
  
      if(head2!=null) {
          current.next = head2;
      }
  
      ListNode result = prehead.next;
      prehead.next = null;
      return result;
  }
  

---