给定一个排序链表，删除所有含有重复数字的节点，只保留原始链表中 没有重复出现 的数字。

示例 1:

输入: 1->2->3->3->4->4->5
输出: 1->2->5

示例 2:

输入: 1->1->1->2->3
输出: 2->3

Solution

• 借助栈保存不重复节点

// @lc code=start
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        stack<ListNode *> stack;
        if (head==nullptr) return head;
        ListNode *cur = head;
        int pre = head->val-1; // 为了确保头结点入栈

        while (cur) {
            if (cur->val != pre) {
                stack.push(cur);
                pre = cur->val;
            } else {
                if (!stack.empty() && stack.top()->val == cur->val) {
                    stack.pop();
                }
            }
            cur = cur->next;
        }

        ListNode *node = nullptr; // 头插法
        while (!stack.empty()) {
            ListNode *tmp = stack.top();
            tmp->next = node;
            node = tmp;
            stack.pop();
        }
        return node;
    }
};
// @lc code=end

参考 liuyubobobo 、@durianyu

• 三指针

// @lc code=start
/**
 * 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* dummyHead = new ListNode(-1, head);
        ListNode* pre = dummyHead, *cur = head;
        while(cur){
            int num = 0;
            ListNode* p = cur;
            while(p && p->val==cur->val){
                p = p->next;
                num += 1;
            }
            if(num>1)
                pre->next = p;
            else 
                pre = pre->next;
            cur = p;
        }
        return dummyHead->next;
    }
};
// @lc code=end