LeetCode 82. Remove Duplicates from Sorted List II

原题链接在这里：https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/

题目：

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

For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

题解：

类似Remove Duplicates from Sorted List, 不同就是要完全去掉duplicate.

当cur.next.val == cur.next.next.val 时，就一直移动dup = cur.next.next直到cur.next.val != dup.val. 然后cur的next指向dup.

Time Complexity: O(n), n 是list长度.

Space: O(1).

AC Java:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if(head == null || head.next == null){
            return head;
        }
        
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode cur = dummy;
        while(cur.next != null && cur.next.next != null){
            if(cur.next.val != cur.next.next.val){
                cur = cur.next;
            }else{
                ListNode dup = cur.next.next;
                while(dup != null && cur.next.val == dup.val){
                    dup = dup.next;
                }
                cur.next = dup;
            }
        }
        
        return dummy.next;
    }
}

AC C++:

/**
 * 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 || !head->next){
            return head;
        }

        ListNode* dummy = new ListNode(-1, head);
        ListNode* it = dummy;
        while(it->next && it->next->next){
            if(it->next->val != it->next->next->val){
                it = it->next;
            }else{
                ListNode* dup = it->next->next;
                while(dup && dup->val == it->next->next->val){
                    dup = dup->next;
                }

                it->next = dup;
            }
        }

        return dummy->next;
    }
};