leetcode 82. Remove Duplicates from Sorted List II

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

Example 1:

Input: 1->2->3->3->4->4->5
Output: 1->2->5

Example 2:

Input: 1->1->1->2->3
Output: 2->3

题目大意：给定已排序的链表，删除所有有重复数字的结点。

c++:

// /**
//  * 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) {
        if ((head == nullptr) || (head->next == nullptr)) return head; //如果链表为空或者只有一个结点，直接返回
        ListNode *ans = new ListNode(0); //为方便后续操作，建立头结点
        ListNode *cur = ans; //cur指针指向结果链表的表尾
        ListNode *slow = head, *fast;
        while (slow != nullptr) {
            fast = slow->next;
            while (fast != nullptr && (fast->val == slow->val)) //只要fast不为空，并且其值和slow->val相等，那么fast往后移
                fast = fast->next;
            if (slow->next == fast) { //slow指向fast，说明slow和fast相邻且值不相等，那么slow指向的结点可以并入结果链表中
                cur->next = slow;
                cur = slow;
                cur->next = nullptr; //很关键，需要及时将结果链表和slow指向的head链表断开。
            }
            slow = fast; //不管前面slow是否指向fast，fast指向的结点都是下一个并入结果链表的候选结点。
            //cur->next = nullptr;
        }
        return ans->next;
    }
};