Given the head of a sorted linked list, delete all nodes that have duplicate numbers,
leaving only distinct numbers from the original list. Return the linked list sorted as well.
Test Cases
Example 1:
Input: head = [1,2,3,3,4,4,5]
Output: [1,2,5]
Example 2:
Input: head = [1,1,1,2,3]
Output: [2,3]
Solution
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* deleteDuplicates(ListNode* head) {
ListNode* prev = new ListNode(0, head);
ListNode* curr = prev;
while(head != nullptr) {
if (head->next != nullptr && head->val == head->next->val) {
while(head->next != nullptr && head->val == head->next->val) {
head = head->next;
}
curr->next = head->next;
} else {
curr = curr->next;
}
head = head->next;
}
return prev->next;
}
};/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode deleteDuplicates(ListNode head) {
ListNode prev = new ListNode(0, head);
ListNode curr = prev;
while(head != null) {
if (head.next != null && head.val == head.next.val) {
while(head.next != null && head.val == head.next.val) {
head = head.next;
}
curr.next = head.next;
} else {
curr = curr.next;
}
head = head.next;
}
return prev.next;
}
}