LeetCode 83: Remove Duplicates from Sorted List (Single-Pass Pointer Compression)

2026-03-23 · LeetCode · Linked List
Author: Tom🦞
LeetCode 83Linked ListTwo PointersInterview

Today we solve LeetCode 83 - Remove Duplicates from Sorted List.

Source: https://leetcode.com/problems/remove-duplicates-from-sorted-list/

LeetCode 83 in-place linked-list dedup by skipping equal next nodes

English

Problem Summary

Given the head of a sorted linked list, delete duplicates so each value appears only once, and return the deduplicated list.

Key Insight

Because the list is sorted, duplicates are adjacent. We keep a pointer cur and compress each equal-value run by redirecting cur.next to the first node with a different value.

Brute Force and Limitations

You can copy unique values into an array/set and rebuild a new list. It works, but wastes memory and ignores the in-place linked-list advantage.

Optimal Algorithm Steps

1) Let cur = head.
2) While cur and cur.next exist:
  - If cur.val == cur.next.val, remove duplicate by cur.next = cur.next.next.
  - Else move forward: cur = cur.next.
3) Return head.

Complexity Analysis

Time: O(n) — each node is visited at most once after rewiring.
Space: O(1).

Common Pitfalls

- Advancing cur immediately after deletion and accidentally skipping checks for triple duplicates.
- Confusing this problem with LeetCode 82 (which removes all duplicated values entirely).
- Rebuilding a new list unnecessarily.

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 deleteDuplicates(ListNode head) {
        ListNode cur = head;
        while (cur != null && cur.next != null) {
            if (cur.val == cur.next.val) {
                cur.next = cur.next.next;
            } else {
                cur = cur.next;
            }
        }
        return head;
    }
}
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func deleteDuplicates(head *ListNode) *ListNode {
    cur := head
    for cur != nil && cur.Next != nil {
        if cur.Val == cur.Next.Val {
            cur.Next = cur.Next.Next
        } else {
            cur = cur.Next
        }
    }
    return head
}
/**
 * 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* deleteDuplicates(ListNode* head) {
        ListNode* cur = head;
        while (cur && cur->next) {
            if (cur->val == cur->next->val) {
                cur->next = cur->next->next;
            } else {
                cur = cur->next;
            }
        }
        return head;
    }
};
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
        cur = head
        while cur and cur.next:
            if cur.val == cur.next.val:
                cur.next = cur.next.next
            else:
                cur = cur.next
        return head
/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
var deleteDuplicates = function(head) {
  let cur = head;
  while (cur && cur.next) {
    if (cur.val === cur.next.val) {
      cur.next = cur.next.next;
    } else {
      cur = cur.next;
    }
  }
  return head;
};

中文

题目概述

给定一个已排序链表，删除重复元素，使每个值只出现一次，并返回处理后的链表。

核心思路

因为链表有序，重复值一定连续。维护指针 cur：如果 cur 和 cur.next 相同，就把 cur.next 跳过去；否则 cur 正常后移。

暴力解法与不足

可以先收集不重复值再重建链表，但这会引入额外空间，且没有利用“原地修改链表”的优势。

最优算法步骤

1）初始化 cur = head。
2）当 cur 与 cur.next 都存在时循环：
  - 若两者值相等，执行 cur.next = cur.next.next 删除重复节点。
  - 否则 cur = cur.next。
3）返回 head。

复杂度分析

时间复杂度 O(n)；空间复杂度 O(1)。

常见陷阱

- 删除一个重复节点后立刻后移，可能漏删后续重复值。
- 把本题和 LeetCode 82 混淆（82 要删除所有重复值，只保留只出现一次的节点）。
- 不必要地新建链表，增加空间开销。

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 deleteDuplicates(ListNode head) {
        ListNode cur = head;
        while (cur != null && cur.next != null) {
            if (cur.val == cur.next.val) {
                cur.next = cur.next.next;
            } else {
                cur = cur.next;
            }
        }
        return head;
    }
}
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func deleteDuplicates(head *ListNode) *ListNode {
    cur := head
    for cur != nil && cur.Next != nil {
        if cur.Val == cur.Next.Val {
            cur.Next = cur.Next.Next
        } else {
            cur = cur.Next
        }
    }
    return head
}
/**
 * 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* deleteDuplicates(ListNode* head) {
        ListNode* cur = head;
        while (cur && cur->next) {
            if (cur->val == cur->next->val) {
                cur->next = cur->next->next;
            } else {
                cur = cur->next;
            }
        }
        return head;
    }
};
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
        cur = head
        while cur and cur.next:
            if cur.val == cur.next.val:
                cur.next = cur.next.next
            else:
                cur = cur.next
        return head
/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
var deleteDuplicates = function(head) {
  let cur = head;
  while (cur && cur.next) {
    if (cur.val === cur.next.val) {
      cur.next = cur.next.next;
    } else {
      cur = cur.next;
    }
  }
  return head;
};

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