LeetCode | Remove Duplicates from Sorted List II

题目：

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.

思路：

思路1：

基本思路是先遍历确定重复的数值，然后再遍历一次构造输出的链表。可以采用Hash表结构，每个Hash值需要记录出现的次数；当然也可以采用两个bitmap，一个用来保存出现一次的数值，另一个用来保存出现两次的数值。

思路2：

一次遍历删除所有重复的结点。

代码：

思路1：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */ class bitmap{public:    bitmap(int n);~bitmap();void set(int k);bool val(int k);private:int len;unsigned int* map;unsigned int* reversemap;bool zero;};bitmap::bitmap(int n){    len = n/32+1;map = new unsigned int[len];reversemap = new unsigned int[len];for(int i = 0; i < len; i++){map[i] = 0;reversemap[i] = 0;zero = false;}}void bitmap::set(int k){bool reverse = true;if(k== 0){zero = true;}else if(k<0){reverse = false;k =-k;}int a = k/32;int b = k%32;if(a >= len){return;}else{if(reverse){int tmp = reversemap[a]/pow(2,b);if(tmp%2 == 0){reversemap[a] += pow(2,b);}}else{int tmp = map[a]/pow(2,b);if(tmp%2 == 0){map[a] += pow(2,b);}}}}bool bitmap::val(int k){bool reverse = true;if(k== 0){return zero;}else if(k<0){reverse = false;k =-k;}int a = k/32;int b = k%32;if(a >= len){return false;}else{if(reverse){int tmp = reversemap[a]/pow(2,b);if(tmp%2 == 1){return true;}else{return false;}}else{int tmp = map[a]/pow(2,b);if(tmp%2 == 1){return true;}else{return false;}}}}class Solution {public:    ListNode *deleteDuplicates(ListNode *head) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if(head == NULL)        {            return NULL;        }                ListNode * p =head;        bitmap* bm = new bitmap(32767);                bitmap* duplicate = new bitmap(32767);                while(p != NULL)        {            if(bm->val(p->val))            {                duplicate->set(p->val);            }            else            {                bm->set(p->val);            }            p = p->next;        };                ListNode * begin = NULL;        p =head;        while(p != NULL)        {            if(!duplicate->val(p->val))            {                begin = p;                break;            }            p = p->next;        };                if(p == NULL)        {            return begin;        }                while(p->next != NULL)        {            if(duplicate->val(p->next->val))            {                p->next = p->next->next;            }            else            {                p = p->next;            }        };        return begin;    }};

思路2：

/** * 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 * ptr = new ListNode(INT_MIN);        ptr->next = head;                ListNode * lead = ptr;                int cur = INT_MIN;                while(ptr->next != NULL && ptr->next->next != NULL){            if(ptr->next->val == ptr->next->next->val){                cur = ptr->next->val;                ptr->next->next = ptr->next->next->next;            }            else{                if(ptr->next->val == cur){                    ptr-> next = ptr->next->next;                }                else{                    ptr = ptr->next;                }            }        }                if(ptr->next != NULL && ptr->next->val == cur){            ptr->next = NULL;        }                        return lead->next;    }};