82. Remove Duplicates from Sorted List II
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.
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]
Constraints:
- The number of nodes in the list is in the range [0, 300].
- -100 <= Node.val <= 100
- The list is guaranteed to be sorted in ascending order.
/**
* 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 dummy = new ListNode(0), fast = head, slow = dummy;
slow.next = fast;
while(fast != null) {
while (fast.next != null && fast.val == fast.next.val) {
//while loop to find the last node of the dups.
fast = fast.next;
}
if (slow.next != fast) { //duplicates detected.
slow.next = fast.next; //remove the dups.
fast = slow.next; //reposition the fast pointer.
}else {
//no dup, move down both pointer.
slow = slow.next;
fast = fast.next;
}
}
return dummy.next;
}
}