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:

这道题跟Remove Duplicates from Sorted List比较类似，只是这里要把出现重复的元素全部删除。其实道理还是一样，只是现在要把前驱指针指向上一个不重复的元素中，如果找到不重复元素，则把前驱指针指向该元素，否则删除此元素。算法只需要一遍扫描，时间复杂度是O(n)，空间只需要几个辅助指针，是O(1)。

一般会改到链表头的题目就会需要一个辅助指针, 比如这里的newHead.

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