82. 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.

注意边界条件和循环的判断以及if-else的条件

添加dummy，简化情况。

/** * 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)            return NULL;        ListNode *dummy = new ListNode(INT_MAX);        dummy->next = head;        ListNode *prev = dummy;        ListNode *cur = head;        ListNode *next = cur->next;        while(cur)        {            if(next && cur->val != next->val)            {                prev->next = cur;                prev = cur;                cur = next;                if(next)                    next = next->next;            }            else if(next && cur->val == next->val)            {                while(next && next->val == cur->val)                {                    ListNode *nextNext = next->next;                    delete next;                    next = nextNext;                }                                delete cur;                cur = next;                prev->next = cur;                if(next)                    next = next->next;            }            else            {                cur = next;                if(next)                    next = next->next;            }        }                return dummy->next;            }};

Remove Duplicates from Sorted List

题意要把链表中有重复的元素全部去除，只留下没有重复过的元素。

依旧是链表的操作，用pre和now去删除节点，由于头节点也有可能删除，因此先建一个Head节点连头节点。

pre从Head开始，用now去判断是否重复。

如果重复，循环到下一个不同元素之前，把之间的相同元素直接跳过。

再向前移动pre和now指针。

/** * 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) {        ListNode *pre,*now,*Head;        if(!head||!head->next)            return head;        Head=new ListNode(-1);        Head->next=head;        pre=Head;        now=head;        while(now&&now->next)        {            if(now->val == now->next->val)            {                while(now->next && now->val == now->next->val)                {                    now=now->next;                }                pre->next=now->next;                now=now->next;            }            else             {                pre=now;                now=now->next;            }        }        head=Head->next;        delete(Head);        return head;    }};