链表基础之翻转，删除

翻转链表

给出一个链表1->2->3->null，这个翻转后的链表为3->2->1->null

class Solution {public:    /**     * @param head: The first node of linked list.     * @return: The new head of reversed linked list.     */    ListNode *reverse(ListNode *head) {        // write your code here        ListNode* pre = NULL;        while(head != NULL){            ListNode* temp = head ->next;            head ->next = pre;            pre = head;            head = temp;        }        return pre;    }};

链表删除重复的结点，留下唯一一个相同结点

Given a sorted linked list, delete all duplicates such that each element appear only once.
For example,
Given1->1->2, return1->2.
Given1->1->2->3->3, return1->2->3.

class Solution {public:    ListNode *deleteDuplicates(ListNode *head) {        if (head == NULL) return head;        ListNode* pHead = head;        while(pHead != NULL){            while(pHead ->next != NULL && pHead ->val == pHead ->next ->val){                pHead ->next = pHead ->next ->next;            }            pHead = pHead ->next;        }        return head;    }};

删除全部相同的链表结点

• 这个在牛客上超时了。。。如有看出哪里的问题麻烦指教

class Solution {public:    ListNode *deleteDuplicates(ListNode *head) {        if (head == NULL || head ->next == NULL) return head;        ListNode* dummy = new ListNode(0);        dummy->next = head;        head = dummy; //将头指针前移        while(head ->next != NULL && head ->next ->next != NULL){            if(head ->next ->val == head ->next ->next ->val){//如果两个值相等，记下这个值，删掉所有相同值的结点                int value = head ->next ->val;                while(head->next != NULL && head ->val == value){                    head ->next = head ->next ->next;                }            }else{                head = head ->next;                }        }        return dummy ->next;    }};