Remove Duplicates from Sorted List II @LeetCode

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.

	package leetcode.linkedlist;
	/**
	* Solution: Use Dummy Node to handle the head change.
	*
	* Difference between II and I is the head will not change even duplicates in I, but will change in II(remove all same).
	* So we need dummy node to delete head node, In addition, we need to remember the value of duplicate node.
	* Usually, only head changes, we use dummy node. We can't change dummy in progress, because it points to head we will return at last.
	*
	* Take care:
	* int val = head.next.val must use while delete this value at once.
	* head = head.next must be wrapped by else, or it may head.next == null, and next loop. null.next leads NullPointer Problem.
	* This is not same as removeduplicate I, while condition is not same.
	*
	*
	* @author jeffwan
	* @date Feb 16, 2014
	*/
	public class DeleteDuplicates2 {
	public ListNode deleteDuplicates(ListNode head) {
	if (head == null || head.next == null) {
	return head;
	}
	ListNode dummy = new ListNode(0);
	dummy.next = head;
	head = dummy;
	while (head.next != null && head.next.next != null) {
	if (head.next.val == head.next.next.val) {
	int val = head.next.val;
	while (head.next != null && head.next.val == val) {
	head.next = head.next.next;
	}
	} else {
	head = head.next;
	}
	}
	return dummy.next;
	}
	class ListNode {
	int val;
	ListNode next;
	ListNode (int x) {
	this.val = x;
	}
	}
	}

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