---
tags: leetcode
---

# [82. Remove Duplicates from Sorted List II](https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/)
---
# My Solution
## The Key Idea for Solving This Coding Question
## C++ Code

/**
 * 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) {
        if (head == nullptr || head->next == nullptr) {
            return head;
        }
        // At least, we have two nodes in the linked list.

        ListNode *dummy = new ListNode();
        ListNode *prev = dummy, *left = head, *right = head;
        dummy->next = head;

        while (right != nullptr) {
            if (left->val == right->val) {
                right = right->next;
                continue;
            }

            if (left->next == right) {
                right = right->next;
                left = left->next;
                prev = prev->next;
                continue;
            }

            deleteNodes(prev, left, right);
            left = right;
        }

        if (left->next != right) {
            deleteNodes(prev, left, right);
        }

        head = dummy->next;
        delete dummy;
        return head;
    }

private:
    void deleteNodes(ListNode *prev, ListNode *left, ListNode *right) {
        prev->next = right;
        while (left != right) {
            ListNode *x = left;
            left = left->next;
            delete x;
        }
    }
};

## Time Complexity
$O(n)$
$n$ is the number of nodes in the linked list referred by `head`.
## Space Complexity
$O(1)$
## C++ Code 2

/**
 * 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 *dummy = new ListNode(-200, head), *prev = dummy, *curr = head;
        
        while (curr && curr->next) {
            if (curr->val == curr->next->val) {
                curr = curr->next;
                continue;
            }
            
            if (prev->next == curr) {
                prev = prev->next;
                curr = curr->next;
            } else {
                curr = curr->next;
                deleteDuplicates(prev, curr);
            }
        }
        
        if (prev->next != curr) {
            curr = curr->next;
            deleteDuplicates(prev, curr);
        }
        
        head = dummy->next;
        delete dummy;
        
        return head;
    }
    
private:
    void deleteDuplicates(ListNode *left, ListNode *right) {
        while (left->next != right) {
            ListNode *x = left->next;
            left->next = x->next;
            delete x;
        }
    }
};

# Miscellaneous
<!--
# Test Cases

[1,2,3,3,4,4,5]

[1,1,1,2,3]

[1,2,3,3,3,3]

[1,2,3,3,4,5,5,5,6,7]

[1,1,1,2,3,3,3,4,5,5,5]

[]

[2]

[1,1,1,1,1,1,1]

-->