题目

给定一个排序链表，删除所有重复的元素，使得每个元素只出现一次。

示例 1:

输入: 1->1->2
输出: 1->2
示例 2:

输入: 1->1->2->3->3
输出: 1->2->3

解析

由于已经是排序链表，直接判断next的值是不是等于当前节点的值

代码

/**
 * 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* cur = head;
        while (cur && cur->next) {
            if (cur->val == cur->next->val) cur->next = cur->next->next;
            else cur = cur->next;
        }
        return head;
    }
};

如果链表是无序的，用unordered_set记录即可

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
// private:
    // unordered_set<int> record;
public:
    ListNode* deleteDuplicates(ListNode* head) {
        // ListNode* h = head;
        ListNode* dummy = new ListNode(-1);
        dummy->next = head;
        ListNode* pre = dummy;
        ListNode* cur = dummy->next;
        while (cur) {
            if (record.find(cur->val) == record.end()) {
                record.insert(cur->val);
                pre = pre->next;
                cur = cur->next;
            }
            else {
                pre->next = cur->next;
                cur = cur->next;
            }
        }
        return dummy->next;
    }
};