Remove Linked List Elements &&remove-duplicates-from-sorted-list

Remove all elements from a linked list of integers that have value val.

样例

Given 1->2->3->3->4->5->3, val = 3, you should return the list as 1->2->4->5

这道题的算法其实很简单，如果遇到要删除的元素，跳过，直接找到下一个非目标元素；无奈学渣的我对链表竟然已经到了如此退化的题目，所以也研究了一些时间。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    /**
     * @param head a ListNode
     * @param val an integer
     * @return a ListNode
     */
    public ListNode removeElements(ListNode head, int val) {
        // Write your code here
        ListNode helpnode = new ListNode (0);
        helpnode.next = head;
        ListNode p = helpnode;
        while(p.next != null){
            if(p.next.val == val){
                p.next = p.next.next;
            }
            else {
                p = p.next;
            }
        }
        return helpnode.next;
    }
}

拓展题：

删除排序链表中的重复元素

容易 删除排序链表中的重复元素查看运行结果 

给定一个排序链表，删除所有重复的元素每个元素只留下一个

样例

给出1->1->2->null，返回 1->2->null

给出1->1->2->3->3->null，返回 1->2->3->null

/**
 * 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 linked list
     */
    public static ListNode deleteDuplicates(ListNode head) { 
        ListNode p = new ListNode(0);
        p.next = head;
        if(head == null )
           return null;
        while (head.next != null){
            if (head.next.val == head.val ){
                 head.next = head.next.next;
             }else{
             head = head.next;
             }
        }
        return  p.next;
    }  
}