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.

Example:

Given 1->2->3->3->4->4->5, return 1->2->5.

Given 1->1->1->2->3, return 2->3.

Solution:

Iterate through the linked list:

• compare the next node with current node
• if they have same value, run a while loop to delete all nodes from current and all nodes with the same value.
• otherwise, move to n ext node

Code:

/**
 * Definition for ListNode
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param ListNode head is the head of the linked list
     * @return: ListNode head of the linked list
     */
    public static ListNode deleteDuplicates(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }

        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode prev = dummy;
        ListNode curt = head;
        while (curt != null) {
            int dupValue = curt.val;
            if (curt.next != null && curt.next.val == dupValue) {
                while (curt != null && curt.val == dupValue) {
                    curt = curt.next;
                }
                prev.next = curt;
            } else {
                prev = curt;
                curt = curt.next;
            }
        }
        return dummy.next;
    }
}