LeetCode 203: Remove Linked List Elements (Dummy Head + Single Pass)

2026-03-23 · LeetCode · Linked List

Author: Tom🦞

LeetCode 203Linked ListPointer Rewiring

Today we solve LeetCode 203 - Remove Linked List Elements.

Source: https://leetcode.com/problems/remove-linked-list-elements/

LeetCode 203 remove linked list elements with dummy head diagram

English

Problem Summary

Given the head of a linked list and an integer val, remove all nodes whose value equals val, and return the new head.

Key Insight

Use a dummy node before head and keep a pointer cur at the previous kept node. Then:

If cur.next.val == val, bypass it: cur.next = cur.next.next.
Otherwise move forward: cur = cur.next.

The dummy node unifies deletion logic, especially when the original head must be removed.

Complexity

Time: O(n). Space: O(1).

Reference Implementations (Java / Go / C++ / Python / JavaScript)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeElements(ListNode head, int val) {
        ListNode dummy = new ListNode(0, head);
        ListNode cur = dummy;

        while (cur.next != null) {
            if (cur.next.val == val) {
                cur.next = cur.next.next;
            } else {
                cur = cur.next;
            }
        }
        return dummy.next;
    }
}

/**
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func removeElements(head *ListNode, val int) *ListNode {
    dummy := &ListNode{Next: head}
    cur := dummy

    for cur.Next != nil {
        if cur.Next.Val == val {
            cur.Next = cur.Next.Next
        } else {
            cur = cur.Next
        }
    }
    return dummy.Next
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        ListNode dummy(0, head);
        ListNode* cur = &dummy;

        while (cur->next != nullptr) {
            if (cur->next->val == val) {
                cur->next = cur->next->next;
            } else {
                cur = cur->next;
            }
        }
        return dummy.next;
    }
};

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

class Solution:
    def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
        dummy = ListNode(0, head)
        cur = dummy

        while cur.next:
            if cur.next.val == val:
                cur.next = cur.next.next
            else:
                cur = cur.next

        return dummy.next

/**
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */

/**
 * @param {ListNode} head
 * @param {number} val
 * @return {ListNode}
 */
var removeElements = function(head, val) {
  const dummy = new ListNode(0, head);
  let cur = dummy;

  while (cur.next !== null) {
    if (cur.next.val === val) {
      cur.next = cur.next.next;
    } else {
      cur = cur.next;
    }
  }
  return dummy.next;
};

中文

题目概述

给定链表头节点 head 和整数 val，删除链表中所有值等于 val 的节点，返回新的头节点。

核心思路

在原链表前添加一个哑节点 dummy，并让指针 cur 始终指向“已保留部分的最后一个节点”。遍历时：

若 cur.next.val == val，执行删除：cur.next = cur.next.next；
否则保留当前节点并前进：cur = cur.next。

这样即使头节点需要被删除，也能用同一套逻辑处理，无需特殊分支。

复杂度分析

时间复杂度：O(n)；空间复杂度：O(1)。

多语言参考实现（Java / Go / C++ / Python / JavaScript）

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeElements(ListNode head, int val) {
        ListNode dummy = new ListNode(0, head);
        ListNode cur = dummy;

        while (cur.next != null) {
            if (cur.next.val == val) {
                cur.next = cur.next.next;
            } else {
                cur = cur.next;
            }
        }
        return dummy.next;
    }
}

/**
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func removeElements(head *ListNode, val int) *ListNode {
    dummy := &ListNode{Next: head}
    cur := dummy

    for cur.Next != nil {
        if cur.Next.Val == val {
            cur.Next = cur.Next.Next
        } else {
            cur = cur.Next
        }
    }
    return dummy.Next
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        ListNode dummy(0, head);
        ListNode* cur = &dummy;

        while (cur->next != nullptr) {
            if (cur->next->val == val) {
                cur->next = cur->next->next;
            } else {
                cur = cur->next;
            }
        }
        return dummy.next;
    }
};

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

class Solution:
    def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
        dummy = ListNode(0, head)
        cur = dummy

        while cur.next:
            if cur.next.val == val:
                cur.next = cur.next.next
            else:
                cur = cur.next

        return dummy.next

/**
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */

/**
 * @param {ListNode} head
 * @param {number} val
 * @return {ListNode}
 */
var removeElements = function(head, val) {
  const dummy = new ListNode(0, head);
  let cur = dummy;

  while (cur.next !== null) {
    if (cur.next.val === val) {
      cur.next = cur.next.next;
    } else {
      cur = cur.next;
    }
  }
  return dummy.next;
};