Approach: Dummy + skip runs

def deleteDuplicates(self, head):
    dummy = ListNode(0, head)
    prev = dummy
    cur = head
    while cur:
        while cur.next and cur.val == cur.next.val:
            cur = cur.next
        if prev.next == cur:
            prev = prev.next
        else:
            prev.next = cur.next
        cur = cur.next
    return dummy.next

Complexity: O(n) time, O(1) space.

Full Solution — all languages

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next
class Solution:
    def deleteDuplicates(self, head):
        dummy = ListNode(0, head)
        prev = dummy
        cur = head
        while cur:
            while cur.next and cur.val == cur.next.val:
                cur = cur.next
            if prev.next == cur:
                prev = prev.next
            else:
                prev.next = cur.next
            cur = cur.next
        return dummy.next

import java.util.*;
class ListNode {
    int val;
    ListNode next;
    ListNode(int x) {
        val = x;
    }
}
class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode prev = dummy, cur = head;
        while (cur != null) {
            while (cur.next != null && cur.val == cur.next.val) cur = cur.next;
            if (prev.next == cur) prev = prev.next;
            else prev.next = cur.next;
            cur = cur.next;
        }
        return dummy.next;
    }
}

#include <bits/stdc++.h> using namespace std;
struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x): val(x), next(nullptr) {
    }
};
class Solution {
    public: ListNode* deleteDuplicates(ListNode* head) {
        ListNode dummy(0);
        dummy.next = head;
        ListNode* prev = &dummy;
        ListNode* cur = head;
        while (cur) {
            while (cur->next && cur->val == cur->next->val) cur = cur->next;
            if (prev->next == cur) prev = prev->next;
            else prev->next = cur->next;
            cur = cur->next;
        }
        return dummy.next;
    }
};

/** Node shape: { val, next }. Build/return plain objects or use this ctor. */ function ListNode(val, next) {
    this.val = (val === undefined ? 0 : val);
    this.next = (next === undefined ? null : next);
}
/**
 * @param {ListNode} head
 * @return {ListNode}
 */ var deleteDuplicates = function(head) {
    const dummy = new ListNode(0, head);
    let prev = dummy, cur = head;
    while (cur) {
        while (cur.next && cur.val === cur.next.val) cur = cur.next;
        if (prev.next === cur) prev = prev.next;
        else prev.next = cur.next;
        cur = cur.next;
    }
    return dummy.next;
};

Remove Duplicates from Sorted List II

Example 1

Example 2

Constraints