82. Remove Duplicates from Sorted List II

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:

The idea is to keep two pointers.

One points to the current non-duplicate node.

The other points to the candidate node.

Keep advancing the second pointer to the next non-duplicate node.

Connect these two nodes. (In this way we skip the duplicated nodes)

The time complexity is O(n).


Code:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy;
        ListNode curr = head;
        while (curr != null) {
            while (curr.next != null && curr.val == curr.next.val) {
                curr = curr.next;
            }
            if (pre.next == curr) {
                pre = pre.next;
            }
            else {
                pre.next = curr.next;
            }
            curr = curr.next;
        }
        return dummy.next;
    }
}