Reference: LeetCode
Difficulty: Medium

Problem

Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.

You should try to do it in place. The program should run in $O(1)$ space complexity and $O(nodes)$ time complexity.

Note:

  • The relative order inside both the even and odd groups should remain as it was in the input.
  • The first node is considered odd, the second node even and so on …

Example:

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Input: 1->2->3->4->5->NULL
Output: 1->3->5->2->4->NULL
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Input: 2->1->3->5->6->4->7->NULL
Output: 2->3->6->7->1->5->4->NULL

Analysis

Methods:

  1. OddList & EvenList
    • Use two lists to store two types of nodes and connect them in the end.
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      //  1  2  3  4  5
      //
      // 1 3 5 2 4
      public ListNode oddEvenList(ListNode head) {
      if (head == null || head.next == null) {
      return head;
      }
      ListNode oddDummy = new ListNode(-1), evenDummy = new ListNode(-1);
      ListNode oddPrev = oddDummy, evenPrev = evenDummy;

      int count = 1;
      while (head != null) {
      if (count % 2 != 0) { // odd
      oddPrev.next = head;
      oddPrev = oddPrev.next;
      } else { // even
      evenPrev.next = head;
      evenPrev = evenPrev.next;
      }
      head = head.next;
      count += 1;
      }
      // concatenation
      oddPrev.next = evenDummy.next;
      evenPrev.next = null;
      return oddDummy.next;
      }

Time: $O(N)$
Space: $O(1)$

  1. Solution
    • Do not use count.
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      public ListNode oddEvenList(ListNode head) {
      if (head == null || head.next == null) {
      return head;
      }
      ListNode odd = head, even = head.next, evenHead = even;
      while (even != null && even.next != null) {
      odd.next = odd.next.next;
      even.next = even.next.next;
      odd = odd.next;
      even = even.next;
      }
      odd.next = evenHead;
      return head;
      }

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