### leetcode Question 80: Remove Duplicates from Sorted List II

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.

Analysis:

To solve this problem, one pointer is needed to store the previous element. See example to get a better understanding.

p is the current element (now its value is 2 in e.g.), p2 is the next element after p, for each p, we check the elements after p (3 3 4), and find the non-duplicate position (4), and remove the duplicates(3->3).

Note. To handle the first is duplicate element, we put another node ahead the head.

e.g.
1 2 3 3 4 4 5
go through, 1 and 2,
1->2->3->3->4
p->p2             check if p2.val==p2.next.val
p->     p2        go to next element and check
p-------->      if all duplicates found, let p.next=p2.next, in order to remove the duplicates.

Code:

/**
* 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) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if ((head==NULL) || (head->next == NULL) ){return head;}
ListNode* pre = new ListNode(0);
pre->next = head;
head =pre;
ListNode* p = head;

while(p->next!=NULL){
ListNode *p2 = p->next;
while ((p2->next!=NULL)&&(p2->val==p2->next->val)){
p2=p2->next;
}
if (p2!=p->next){
p->next=p2->next;
}else{
p=p->next;
}
}
return head->next;
}
};


#### 2 comments:

1. In this situation I used DuplicateFilesDeleter for great effect. It searches two or more duplicate files in one or more selected search paths and removes them.

2. Why do i get a segmentation fault with the following code ?
here's my code:

http://ideone.com/0arsZc