203 Remove Linked List Elements

题目

Remove all elements from a linked list of integers that have value val.

Example:

Input: 1->2->6->3->4->5->6, val = 6
Output: 1->2->3->4->5


解法思路

  • 首先要对传入的头节点 head 做判断,是不是不为空?是不是就是和要删除的值相等?如果是的话,那么就要删除头节点;这个动作一直做,直到头节点不再是要删除的节点,或者链表已经被删没了;
  • 如果链表在上述动作中删没了,那么就向外返回 null;
  • 如果链表没有被删没,那么就从头节点的后继节点开始遍历,判断是否应该被删除;
  • 在删除完链表中所有要删除的元素后,返回头节点;

解法实现

时间复杂度
  • O(N);
空间复杂度
  • O(1);
关键字

链表 删除链表中的元素

实现细节
  • 注意对头节点的判断;
package leetcode._203;

public class Solution203_1 {

    public ListNode removeElements(ListNode head, int val) {

        while (head != null && head.val == val) {
            ListNode delNode = head;
            head = delNode.next;
            delNode.next = null;
        }

        if (head == null) {
            return null;
        }

        ListNode prev = head;
        while (prev.next != null) {
            if (prev.next.val == val) {
                ListNode delNode = prev.next;
                prev.next = delNode.next;
                delNode.next = null;
            } else {
                prev = prev.next;
            }
        }
        return head;
    }

}

解法思路(二)

  • 在链表的头节点前链入一个虚拟头节点,这样就无须对头节点的情况做单独的判断和处理;

解法实现(二)

时间复杂度
  • O(N);
空间复杂度
  • O(1);
关键字

虚拟头节点 链表 删除链表中的元素

实现细节
// 203. Remove Linked List Elements
// https://leetcode.com/problems/remove-linked-list-elements/description/
// 使用虚拟头结点
// 时间复杂度: O(n)
// 空间复杂度: O(1)
public class Solution2 {

    public ListNode removeElements(ListNode head, int val) {

        // 创建虚拟头结点
        ListNode dummyHead = new ListNode(0);
        dummyHead.next = head;

        ListNode cur = dummyHead;
        while(cur.next != null){
            if(cur.next.val == val ){
                ListNode delNode = cur.next;
                cur.next = delNode.next;
            }
            else
                cur = cur.next;
        }

        return dummyHead.next;
    }

    public static void main(String[] args) {

        int[] arr = {1, 2, 6, 3, 4, 5, 6};
        int val = 6;

        ListNode head = new ListNode(arr);
        System.out.println(head);

        (new Solution1()).removeElements(head, val);
        System.out.println(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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