algorithm - 如何从单向链表的末尾找到第 n 个元素？

Question

以下函数试图找到单向链表的 nth 到 last 元素。

例如:

如果元素是 8->10->5->7->2->1->5->4->10->10 那么结果是7th 到最后一个节点是 7。

任何人都可以帮助我了解这段代码是如何工作的，或者是否有更好更简单的方法？

LinkedListNode nthToLast(LinkedListNode head, int n) {
  if (head == null || n < 1) {
    return null;
  }

  LinkedListNode p1 = head;
  LinkedListNode p2 = head;

  for (int j = 0; j < n - 1; ++j) { // skip n-1 steps ahead
    if (p2 == null) {
      return null; // not found since list size < n
    }
    p2 = p2.next;
  }

  while (p2.next != null) {
    p1 = p1.next;
    p2 = p2.next;
  }

  return p1;
}

Answer 1

这个算法的关键是设置两个指针 p1 和 p2 最初相隔 n-1 个节点所以我们想要 p2 指向列表开头的 (n-1)th 节点然后我们移动 p2 直到它到达 last 列表的节点。一旦 p2 到达列表的末尾，p1 将指向列表末尾的第 n 个节点。

我已将解释内嵌为注释。希望对您有所帮助:

// Function to return the nth node from the end of a linked list.
// Takes the head pointer to the list and n as input
// Returns the nth node from the end if one exists else returns NULL.
LinkedListNode nthToLast(LinkedListNode head, int n) {
  // If list does not exist or if there are no elements in the list,return NULL
  if (head == null || n < 1) {
    return null;
  }

  // make pointers p1 and p2 point to the start of the list.
  LinkedListNode p1 = head;
  LinkedListNode p2 = head;

  // The key to this algorithm is to set p1 and p2 apart by n-1 nodes initially
  // so we want p2 to point to the (n-1)th node from the start of the list
  // then we move p2 till it reaches the last node of the list. 
  // Once p2 reaches end of the list p1 will be pointing to the nth node 
  // from the end of the list.

  // loop to move p2.
  for (int j = 0; j < n - 1; ++j) { 
   // while moving p2 check if it becomes NULL, that is if it reaches the end
   // of the list. That would mean the list has less than n nodes, so its not 
   // possible to find nth from last, so return NULL.
   if (p2 == null) {
       return null; 
   }
   // move p2 forward.
   p2 = p2.next;
  }

  // at this point p2 is (n-1) nodes ahead of p1. Now keep moving both forward
  // till p2 reaches the last node in the list.
  while (p2.next != null) {
    p1 = p1.next;
    p2 = p2.next;
  }

   // at this point p2 has reached the last node in the list and p1 will be
   // pointing to the nth node from the last..so return it.
   return p1;
 }

或者，我们可以将 p1 和 p2 设置为 n 个节点而不是 (n-1)，然后移动 p2 直到列表的末尾而不是移动到最后一个节点:

LinkedListNode p1 = head;
LinkedListNode p2 = head;
for (int j = 0; j < n ; ++j) { // make then n nodes apart.
    if (p2 == null) {
        return null;
    }
    p2 = p2.next;
}
while (p2 != null) { // move till p2 goes past the end of the list.
    p1 = p1.next;
    p2 = p2.next;
}
return p1;