删除链表中重复的节点

解题思路：

1. 我们每次都判断当前结点的值与下一个节点的值是否重复
2. 如果重复就循环寻找下一个不重复的节点，将他们链接到——新链表的尾部（其实就是删除重复的节点）

 

public ListNode deleteDuplication(ListNode pHead) {
    if (pHead == null || pHead.next == null)
        return pHead;
    ListNode next = pHead.next;
    if (pHead.val == next.val) {
        while (next != null && pHead.val == next.val)
            next = next.next;
        return deleteDuplication(next);
    } else {
        pHead.next = deleteDuplication(pHead.next);
        return pHead;
    }
}

非递归版：

public class Solution {
    public ListNode deleteDuplication(ListNode pHead)
    {
if (pHead==null || pHead.next==null){return pHead;}
ListNode Head = new ListNode(0);
Head.next = pHead;
ListNode pre  = Head;
ListNode last = Head.next;
while (last!=null){
    if(last.next!=null && last.val == last.next.val){
        // 找到最后的一个相同节点
        while (last.next!=null && last.val == last.next.val){
            last = last.next;
        }
        pre.next = last.next;
        last = last.next;
    }else{
        pre = pre.next;
        last = last.next;
    }
}
return Head.next;
}
}

”

来源：https://my.oschina.net/u/3973793/blog/3129154