給定一個排序連結串列，刪除所有重複的元素只留下原連結串列中沒有重複的元素。
樣例
給出 1->2->3->3->4->4->5->null，返回 1->2->5->null
給出 1->1->1->2->3->null，返回 2->3->null

一刷ac
解題思路：設定一個標誌位代表是否存在重複，然後拼接連結串列。

/**
 * 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);
        ListNode node = dummy;
        int
 
 flag = 0;
        while(head != null){
            if(head.next != null){
                if(head.next.val == head.val){
                    head = head.next;
                    flag = 1;
                }else{
                    if(flag == 0){
                        node.next = head;
                        node = node.next;
                        head = head.next;
                    }else
 
{
                        head = head.next;
                        flag = 0;
                    }
                }
            }
            else{
                if(flag == 0){
                    node.next = head;
                    node = node.next;
                    head = head.next;
                }else{
                    node.next = null;
                    head = head.next;
                }
            }
        }
        return dummy.next;
    }
}

簡潔版

/**
 * 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);
        ListNode node = dummy;
        dummy.next = head;
        while(head != null){
            while(head.next != null && head.next.val == head.val) head = head.next;
            if(node.next == head){
                head = head.next;
                node = node.next;
            }else{
                node.next = head.next;
                head = head.next;
            }
        }
        return dummy.next;
    }
}