[LeetCode January Challange] Day 5 - Remove Duplicates from Sorted List II

Given the head of a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. Return the linked list sorted as well.

Example 1:

Input: head = [1,2,3,3,4,4,5]
Output: [1,2,5]

Example 2:

Input: head = [1,1,1,2,3]
Output: [2,3]

Constraints:

• The number of nodes in the list is in the range [0, 300].
• -100 <= Node.val <= 100
• The list is guaranteed to be sorted in ascending order.

---

Solution

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

/**
 * 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_head = new ListNode(0, head);
        ListNode* prev = dummy_head;
        
        while (head) {
            if (head->next && head->val == head->next->val) {
                ListNode* del_start = head;
                while (head->next && head->val == head->next->val)
                    head = head->next;
                ListNode* del_end = head->next;
                
                prev->next = head->next;
                head = prev;
                
                while (del_start != del_end) {
                    ListNode *tmp = del_start;
                    del_start = del_start->next;
                    delete tmp;
                }
            } else
                prev = prev->next;
            
            head = head->next;
        }
        
        return dummy_head->next;
    }
};

邊走邊找目標 node (重覆值出現的node)，找到時記錄其起末位置，供後續刪除用。
一旦找到目標 node ，就一直往下走，直到不是目標 node 為止。
將 prev 與結束位置相連，中間都是要刪掉的。