每个重复的元素只留下一个,通过修改next指针的位置来实现
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    1. /**
    2. * Definition for singly-linked list.
    3. * struct ListNode {
    4. * int val;
    5. * ListNode *next;
    6. * ListNode(int x) : val(x), next(NULL) {}
    7. * };
    8. */
    9. class Solution {
    10. public:
    11. ListNode* deleteDuplicates(ListNode* head) {
    12. auto dummy = new ListNode(0);
    13. dummy->next = head;
    14. auto left = head;
    15. while(left)
    16. {
    17. auto right = left->next;
    18. while(right && left->val == right->val)
    19. {
    20. right = right->next;
    21. }
    22. left->next = right;
    23. left = left->next;
    24. }
    25. return dummy->next;
    26. }
    27. };

    第二次写题

    /**
     * Definition for singly-linked list.
     * struct ListNode {
     *     int val;
     *     ListNode *next;
     *     ListNode(int x) : val(x), next(NULL) {}
     * };
     */
    class Solution {
    public:
        ListNode* deleteDuplicates(ListNode* head) {
            ListNode* dummy = new ListNode(0);
            dummy->next = head;
            while(head)
            {
                auto pt = head->next;
                while(pt && pt->val == head->val) pt = pt->next;
                head->next = pt;
                head = head->next;
            }
    
            return dummy->next;
    
        }
    };