82. Remove Duplicates from Sorted List II【Medium】

82. Remove Duplicates from Sorted List II【Medium】

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.

解法一：

 /**
  * 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 == NULL || head->next == NULL) {
             return head;
         }
 
         ListNode * dummy = new ListNode(INT_MIN);
         dummy->next = head;
         head = dummy;
         int duplicate;
 
         while (head->next != NULL && head->next->next != NULL) {
             if (head->next->val == head->next->next->val) {
                 duplicate = head->next->val;
                 while (head->next != NULL && head->next->val == duplicate) {
                     ListNode * temp = head->next;
                     free(temp);
                     head->next = head->next->next;
                 }
             }
             else {
                 head = head->next;
             }
         }
         return dummy->next;
 
     }
 };

nc

解法二：

 class Solution {
 public:
     ListNode* deleteDuplicates(ListNode* head) {
         if(!head) return head;
         ListNode dummy();
         dummy.next = head;
         ListNode *pre = &dummy;
         ListNode *cur = head;
 
         while(cur && cur->next){
             while(cur->next && cur->val == cur->next->val) cur = cur->next;
             if(pre->next == cur){
                 pre = cur;
                 cur = cur->next;
             } else {
                 cur = cur->next;
                 pre->next = cur;
             }
         }
 
         return dummy.next;
     }
 };

参考了@liismn 的代码

解法三：

 class Solution {
 public:
     ListNode* deleteDuplicates(ListNode* head) {
         if(!head||!head->next) return head;
         ListNode* dummy = new ListNode();
         ListNode* tail = dummy;
         int flag = true; // should the current head be added ?
         while(head){
             while(head&&head->next&&head->val==head->next->val)
             {
                 flag = false; // finds duplicate, set it to false
                 head = head->next;
             }
             if(flag) // if should be added
             {
                 tail->next = head;
                 tail = tail->next;
             }
             head = head->next;
             flag = true; // time for a new head value, set flag back to true
         }
         tail->next = nullptr; // Don't forget this... I did..
         return dummy->next;
     }
 };

参考了@GoGoDong 的代码

解法四：

 public ListNode deleteDuplicates(ListNode head) {
     if (head == null) return null;
 
     if (head.next != null && head.val == head.next.val) {
         while (head.next != null && head.val == head.next.val) {
             head = head.next;
         }
         return deleteDuplicates(head.next);
     } else {
         head.next = deleteDuplicates(head.next);
     }
     return head;
 }

递归，参考了@totalheap 的代码，还不太明白