Problem
Solution
/**
* 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 current = head;
while (current != null && current.next != null) {
if (current.val == current.next.val) {
current.next = current.next.next;
} else {
current = current.next;
}
}
return head;
}
}
- Two Pointers 의 대표적인 문제
- current 를 기준으로 보되, current.next 를 같이 보는 것이 주된 포인트!
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