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.

分析：和那一题一样的思想

/** * 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) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        if(head == NULL || head->next == NULL)return head;        ListNode *index = head, *p = head->next, *pre = head;        while(p != NULL)        {            if(p->val != pre->val)            {                index = index->next;                index->val = p->val;            }            pre = p;            p = p->next;        }        p = index->next;        index->next = NULL;        while(p != NULL)        {
    //销毁多余的节点            ListNode *tmp = p;            p = p->next;            delete tmp;        }        return head;    }};

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

For example,

Given 1->2->3->3->4->4->5, return 1->2->5.

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

/** * 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) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        if(head == NULL)return head;        //为操作方便添加头结点        ListNode *addHead = new ListNode(0);        addHead->next = head;        //index 指向已经保存的最后一个节点        ListNode *index = addHead, *p = head;        while(p != NULL)        {            if(p->next == NULL || p->val != p->next->val)            {
    //当前节点没有重复                index = index->next;                index->val = p->val;                p = p->next;            }            else            {
    //当前节点有重复，找到当前节点的最后一个副本的下一个元素                while(p->next && p->val == p->next->val)                    p = p->next;                p = p->next;            }        }        p = index->next;        index->next = NULL;        while(p != NULL)        {
    //销毁多余的节点            ListNode *tmp = p;            p = p->next;            delete tmp;        }        head = addHead->next;        delete addHead;        return head;    }};

 

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