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24. Swap Nodes in Pairs 👍

  • Time: $O(n)$
  • Space: $O(1)$
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class Solution {
 public:
  ListNode* swapPairs(ListNode* head) {
    const int length = getLength(head);
    ListNode dummy(0, head);
    ListNode* prev = &dummy;
    ListNode* curr = head;

    for (int i = 0; i < length / 2; ++i) {
      ListNode* next = curr->next;
      curr->next = next->next;
      next->next = prev->next;
      prev->next = next;
      prev = curr;
      curr = curr->next;
    }

    return dummy.next;
  }

 private:
  int getLength(ListNode* head) {
    int length = 0;
    for (ListNode* curr = head; curr; curr = curr->next)
      ++length;
    return length;
  }
};

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class Solution {
  public ListNode swapPairs(ListNode head) {
    final int length = getLength(head);
    ListNode dummy = new ListNode(0, head);
    ListNode prev = dummy;
    ListNode curr = head;

    for (int i = 0; i < length / 2; ++i) {
      ListNode next = curr.next;
      curr.next = next.next;
      next.next = curr;
      prev.next = next;
      prev = curr;
      curr = curr.next;
    }

    return dummy.next;
  }

  private int getLength(ListNode head) {
    int length = 0;
    for (ListNode curr = head; curr != null; curr = curr.next)
      ++length;
    return length;
  }
}

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class Solution:
  def swapPairs(self, head: ListNode) -> ListNode:
    def getLength(head: ListNode) -> int:
      length = 0
      while head:
        length += 1
        head = head.next
      return length

    length = getLength(head)
    dummy = ListNode(0, head)
    prev = dummy
    curr = head

    for _ in range(length // 2):
      next = curr.next
      curr.next = next.next
      next.next = prev.next
      prev.next = next
      prev = curr
      curr = curr.next

    return dummy.next