LeetCode 之 Remove Duplicates from Sorted List I II — C++ 实现

Remove Duplicates from Sorted List

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.

给定一个有序链表，删除所有多余的元素，使链表中的每个元素只出现一次。

例如：

给定链表 1->1->2，返回 1->2

给定链表 1->1->2->3->3，返回1->2->3。

class Solution {public:    ListNode* deleteDuplicates(ListNode* head) {        if(!head)        {            return NULL;        }                ListNode *pre = head, *p = head->next, *cur = NULL;        int lastVal = head->val;                while(p)        {            if(p->val == lastVal)//重复，则删除            {                cur = p;                p = p->next;                pre->next = p;                delete cur;                cur = NULL;            }            else//不重复            {                lastVal = p->val;                p = p->next;                pre = pre->next;            }        }                return head;    }};

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.

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

给定一个有序链表，删除所有重复元素，只留下不重复的元素。

例如：

给定链表 1->2->3->3->4->4->5，返回 1->2->5

给定链表 1->1->1->2->3，返回 2->3。

分析：

    从表头开始扫描，遇到重复的元素只删除当前的一个，再下次遇到不重复的元素时将这个最开始重复的元素也删掉即可。注意一直重复到表尾的情况。

class Solution {public:    ListNode* deleteDuplicates(ListNode* head) {        if(NULL == head)// 空        {            return NULL;        }                struct ListNode *cur = NULL, *ne = NULL;        int preVal = 0;        int isequal = 0;                ListNode *tempHead = new ListNode(0), *pre = tempHead;//新建临时头结点，不用判断是否删除为头        tempHead->next = head;        ne = head->next;        preVal = head->val;                while(ne)        {            if(ne->val == preVal)//和前一个节点值相等            {                isequal = 1;                cur = ne;                ne = ne->next;                pre->next->next = ne;//pre始终指向当前节点的前两个节点，因为当先节点可能和后面节点重复                                delete cur;//删除重复节点                cur = NULL;            }            else            {                if(isequal)//前面出现过相等的节点，第一个相等的节点没有删除                {                    isequal = 0; //标志清 0                    cur = pre->next;                    pre->next = ne;                                        delete cur;                    cur = NULL;                }                else                {                    pre = pre->next;                }                preVal = ne->val;//从新换成当前节点的新值                ne = ne->next;            }        }        if(isequal)//前面出现过相等的节点，第一个相等的节点没有删除        {            cur = pre->next;            pre->next = NULL;                        delete cur;            cur = NULL;        }                head = tempHead->next;        delete tempHead;                return head;    }};