从单链表中删除元素 (C++)

Posted 2023-02-22

【中文标题】从单链表中删除元素 (C++)

【英文标题】：Removing elements from singly-linked list (C++)

【发布时间】：2017-10-21 23:10:20

【问题描述】：

给定一个由整数 l 和一个整数 k 组成的单链表，如何从列表 l 中删除所有值为 k 的元素？最好在 O(n) 时间内。这是我所拥有的：

// Definition for singly-linked list:
// template<typename T>
// struct ListNode 
//   ListNode(const T &v) : value(v), next(nullptr) 
//   T value;
//   ListNode *next;
// ;
//
ListNode<int> * removeKFromList(ListNode<int> * l, int k) 
ListNode iterator = l;
while(iterator != NULL)
  
    if(iterator->value == k)
    

    
    iterator = iterator->next;
  

有什么建议吗？

【问题讨论】：

您遇到了什么问题？您发布的代码是一个存根：您是否尝试过实现实际的删除机制？

如果您将iterator = iterator->next; 放在else 中，可能会更容易。如果您只是在if 中删除了iterator 处的节点，则您不希望此运行。

提示 #1：使用 std::list 或 std::forward_list。提示 #2：查找删除/擦除习语。

【参考方案1】：

这个怎么样？

ListNode1<int> * removeKFromList(ListNode1<int> * l, int k) 
          ListNode1<int> *tmp = l,*prev=NULL;
          if (l==NULL) return l;
          while (tmp) 
              if (tmp->value == k)
              
                  if (prev)
                      prev->next=tmp->next;
                  else
                      l=l->next;
                  //delete tmp; ??
              
              else
                  prev=tmp;

              tmp=tmp->next;
          
          return l;
      

【参考方案2】：

def removeKFromList(l, k):
    if(l is None):
        return l
    curr = l
    final = ListNode(0)
    i = 1
    newhead = l
    while(curr):
        if(curr.value != k and i):
            i = 0
            newhead = curr
        if(curr.value == k):
            final.next = curr.next
            tmp = final.next
            if(tmp is None and i):
                return None
        else:
            final = curr
        curr = curr.next
    return newhead

【讨论】：

请为您的回答提供一些背景信息。它将帮助访问该站点的其他人轻松理解代码。

【参考方案3】：

**这里是我解决问题的方法和每一步的解释

# Singly-linked lists are already defined with this interface:
# class ListNode(object):
#   def __init__(self, x):
#     self.value = x
#     self.next = None
#
def solution(l, k):
    
    """ 
    - First, check if the is empty or not, so return the empty list itself
    """
    if l is None:
        return l

    """
    - Then trying to remove the occurrences of element k in the head of the 
    linked list ( It may appear several consecutive k elements at the 
    beginning of the linked list. Even the whole linked list may compose of
    nodes with k values all )
    """
    while l.value == k and l != None:
        l = l.next
        if l == None:
            break
    """
    - Then I am checking again if the list is empty after the previous cleaning
    operation in the while loop, because it may remove all of the nodes from 
    the list ( this is a special case when the whole linked list may compose of 
    nodes with k values all)
    """
    if l is None:
        return l

    """
    - Then I am finally checking all of the remaining nodes in the list 
    and removing the connection of the nodes where the value is equal to k.
    """
    curr = l
    while curr.next:
        if int(curr.next.value == k):
            curr.next = curr.next.next
        else:
            curr = curr.next
        
    return l