Description
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.
Solution
Two-pointers
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode deleteDuplicates(ListNode head) {
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode tail = dummy;
ListNode curr = head;
while (curr != null) {
int count = 1;
curr = curr.next;
while (curr != null && curr.val == tail.next.val) {
++count;
curr = curr.next;
}
if (count == 1) { // only 1 node between tail and curr
tail = tail.next;
} else { // discard nodes between tail and curr
tail.next = curr;
}
}
return dummy.next;
}
}
