学步园

Given a sorted linked list, delete all duplicates such that each element appear only
once.

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

思路：

用两个指针。一个指针不断前进，另一个指针不动。比较两个指针对应结点的值。如果不同，则删除两个指针之间的所有元素。之后不动的指针移动到移动指针的位置，移动指针继续前进。要处理好最后几个元素的值相同的情况。

题解：

/**
 * 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) {
        // empty list
        if (head == nullptr)
            return nullptr;
        
        int current_value;
        ListNode* fwd_iter = head->next;
        ListNode* iter = head;

        current_value = head->val;
        while(fwd_iter != nullptr)
        {
            if (current_value != fwd_iter->val)
            {
                if (fwd_iter != iter->next)
                {
                    ListNode* recycle = iter->next;
                    while(recycle != fwd_iter)
                    {
                        ListNode* tmp = recycle;
                        recycle = recycle->next;
                        delete tmp;
                    };
                }
                iter->next = fwd_iter;
                iter = fwd_iter;
                current_value = iter->val;
            }
            fwd_iter = fwd_iter->next;
        }
        
        if (iter->next != nullptr)
        {
            ListNode* recycle = iter->next;
            while(recycle != nullptr)
            {
                ListNode* tmp = recycle;
                recycle = recycle->next;
                delete tmp;
            }
            iter->next = nullptr;
        }
        
        return head;
    }
};