Problems:

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:

This problem is a bit different from the classic probelm of removing duplicates. Here we need to delete all nodes of duplication. To achieve this, three pointers and a flag are needed.

ListNode *ppre is used to point to the node that before dup;

ListNode *pre is used to point to the first node of the duplication;

ListNode *cur is used to point to the current checking node;

bool isDup is used to indicate whether there's duplication;

The checking process is as follows:

1. first check whther pre.val equals cur.vla or not; if equal go to step 2 else go to step 3

2. remove the cur node, and set the isDup flag; go back to step 1;

3. check whether isDup is set or not; if set, delete the pre node. otherwise move all three pointers one node further.

The space needed is constant and the time complexity is O(n).

 

One thing to remember is that after exitting the while loop, we need an extra check of isDup to decide whether delete or not the last node of the list

 

Code:

1 /** 2  * Definition for singly-linked list. 3  * struct ListNode { 4  *     int val; 5  *     ListNode *next; 6  *     ListNode(int x) : val(x), next(NULL) {} 7  * }; 8  */ 9 class Solution {10 public:11     ListNode *deleteDuplicates(ListNode *head) {12         // Start typing your C/C++ solution below13         // DO NOT write int main() function14         if (head == NULL || head->next == NULL)15             return head;16         17         ListNode dmy(-1);18         dmy.next = head;19         20         ListNode *ppre = &dmy, *pre = head, *cur = head->next;21         bool isDup = false;22         23         while (cur != NULL) {24             if (cur->val == pre->val) {25                 isDup = true;26                 pre->next = cur->next;27                 cur->next = NULL;28                 delete cur;29                 cur = pre->next;30             } else {31                 if (isDup) {32                     ppre->next = pre->next;33                     pre->next = NULL;34                     delete pre;35                     pre = cur;36                     cur = cur->next;37                     isDup = false;38                 } else {39                     ppre = pre;40                     pre = cur;41                     cur = cur->next;42                 }43             }44         }45         46         if (isDup) {47             ppre->next = pre->next;48             pre->next = NULL;49             delete pre;50         }51             52         return dmy.next;53     }54 };