Remove Linked List Elements

本文介绍了一种从单链表中移除特定值的所有节点的方法,并提供了详细的Python实现示例。通过迭代的方式找到并移除所有匹配给定值的节点。

地址:点击打开链接

删除链表中为给定的值的元素的节点

注意判断条件

答案:

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def removeElements(self, head, val):
        """
        :type head: ListNode
        :type val: int
        :rtype: ListNode
        """
        while head and head.val == val:
                head = head.next
        if not head :
            return head
        p = head
        q = head.next
        while q:
            if q.val == val:
                q = q.next
            else: #下面主要是考虑到1,6,6,1这种情况
                p.next = q
                p = p.next
                q = q.next
        p.next = q #考虑到1,2这种情况
        return head


### Remove Algorithm in Programming In programming, the `remove` algorithm typically refers to operations that eliminate specific elements from a collection such as arrays, lists, or other data structures. This operation can be implemented differently depending on the language and structure being used. For instance, when dealing with linked lists, removing an element involves adjusting pointers so that the previous node points to the next one, effectively skipping over the node meant for removal[^4]. Below is how you might implement a remove function within the context of a singly linked list: ```python class Node: def __init__(self, data=None): self.data = data self.next = None class LinkedList: def __init__(self): self.head = None def append(self, data): if not self.head: self.head = Node(data) else: current = self.head while current.next: current = current.next current.next = Node(data) def remove(self, key): temp = self.head # If head node holds the key to be removed if (temp and temp.data == key): self.head = temp.next temp = None return # Search for the key to be removed, keeping track of the previous node prev = None while(temp and temp.data != key): prev = temp temp = temp.next # If key was not found in the linked list if temp is None: return # Unlink the node from the linked list prev.next = temp.next temp = None def print_list(self): current = self.head while(current): print(current.data, end=" ") current = current.next print() ``` When it comes to more complex algorithms like those provided by C++'s Standard Template Library (STL), there exists a specialized version known as `std::remove`. It does not actually erase any elements but instead moves all occurrences of specified values towards the end of the range and returns an iterator pointing to the new logical end of the sequence[^3]. Here’s an example demonstrating its use alongside vectors: ```cpp #include <vector> #include <algorithm> // For std::remove #include <iostream> int main(){ std::vector<int> v {10, 20, 30, 40, 50}; auto new_end = std::remove(v.begin(), v.end(), 30); // Erase unneeded part after 'new_end' v.erase(new_end, v.end()); for(auto elem : v){ std::cout << elem << " "; } } // Output will exclude number 30. ``` This approach separates concerns between moving unwanted items out of sight versus physically shrinking container size which enhances flexibility especially useful during multi-step transformations involving containers before finalizing their states permanently through explicit calls like vector's `.erase()` member function shown above.
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