[LintCode] Remove Duplicates from Sorted List II

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:

很简单的一个follow up，问题就是怎么判断重复，并且如何删除，思路就是，首先是删除题，所以进行判断的指针都应该是curr.next 是否等于 curr.next.next。然后如果发现相等，记录下这个点，然后不断的把curr.next这个点删除，删除的时候curr就在不停的后移，删除，然后知道重复点没了，进行下面操作。

代码：

/**
 * 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){
            return head;
        }
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode cur = dummy;
        while (cur.next != null && cur.next.next != null){
            if (cur.next.val == cur.next.next.val){
                int temp = cur.next.val;
                while (cur.next != null && cur.next.val == temp){
                    cur.next = cur.next.next;
                } 
            } else {
                cur = cur.next;
            }
        }
        return dummy.next;
    }
}