25. Reverse Nodes in k-Group

 

Given the head of a linked list, reverse the nodes of the list k at a time, and return the modified list.

 

本题我们可以借助一个Stack来解题。我们从链表头部开始，向Stack中插入k个节点，然后在按顺序pop出来实际上就是将这k个节点翻转了一遍。另外如果stack中元素个数不到k个，顺序保持不变即可

 

public ListNode reverseKGroup(ListNode head, int k) {
  ListNode dummy = new ListNode(0);
  ListNode cur = dummy;
  Stack<ListNode> stack = new Stack<>();
  while(head != null){
    ListNode first = head;
    while(stack.size() < k && head != null){
      stack.push(head);
      head = head.next;
    }
    if(stack.size() < k) {
      cur.next = first;
      break;
    }
    cur.next = reverse(stack);
    cur = first;
  }
  return dummy.next;
}

private ListNode reverse(Stack<ListNode> stack){
  if(stack.size() == 0) 
    return null;
  ListNode node = stack.pop();
  node.next = reverse(stack);
  return node;
}