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.

 

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思路：双指针，一个指向要处理的node，一个指向处理完的node，比较两个node的val是否相等，注意，不相等是要将处理完的node->next = NULL,破坏原来的link关系。

 

/** * 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 == NULL)                return NULL;            ListNode dummy(-1);            dummy.next = head;            ListNode * p = head;            p = p->next;            while(p)            {                #if 0                cout << "p->val\t" << p->val << endl;                cout << "head->val\t" << head->val << endl;                #endif                if(p->val != head->val)                {                       head->next = p;                    head = p;                }                   p = p->next;                // to break old link                head->next = NULL;            }               return dummy.next;        }   };