Remove Linked List Elements

本文介绍了一种删除链表中特定值节点的方法。通过先移除头部匹配的节点,再遍历链表找到待删节点的前驱节点进行删除操作。

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问题:删除链表中指定元素的节点。

思路:问题开始看比较简单,只需要在遍历链表的同时比较节点的值与给定的值是否相同,然后做出判断是否删除即                可。但是我们知道在删除链表元素的时候要找出其前一个节点,如果链表的第一个节点就是需要被删除的节点                呢?所以必须要跳过这样的节点。

public class Solution {
     
   public ListNode removeElements(ListNode head, int val) {
		if (head == null) {
			return null;
		}
		while(head!=null&&head.val==val){
			head=head.next;
		}
		ListNode p=head;
		while(p!=null &&p.next!=null){
			if (p.next.val==val){
				p.next=p.next.next;
			}else{
				p=p.next;
			}
		}
		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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