删除链表中重复的结点

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题目描述

在一个排序的链表中，存在重复的结点，请删除该链表中重复的结点，重复的结点不保留，返回链表头指针。例如，链表1->2->3->3->4->4->5 处理后为 1->2->5

解法一：递归

/* public class ListNode {    int val;    ListNode next = null;    ListNode(int val) {        this.val = val;    }}*/public class Solution {    public ListNode deleteDuplication(ListNode pHead)    {if(pHead == null || pHead.next == null) {            return pHead;        }        if(pHead.next.val == pHead.val) {            ListNode curNode = pHead.next.next;            int val = pHead.val;            while(curNode != null && curNode.val == val) {                curNode = curNode.next;            }            return deleteDuplication(curNode);        } else {            pHead.next = deleteDuplication(pHead.next);            return pHead;        }    }}

解法二：非递归

/* public class ListNode {    int val;    ListNode next = null;    ListNode(int val) {        this.val = val;    }}*/public class Solution {    public ListNode deleteDuplication(ListNode pHead)    {if(pHead == null || pHead.next == null) {            return pHead;        }        ListNode root = new ListNode(0);        root.next = pHead;//头结点前驱结点        ListNode pre = root;//pre为移动当前节点的前驱节点        ListNode cur = pHead;//当前节点        while(cur != null && cur.next != null) {            //当前节点与下一个节点值相等            if(cur.next.val == cur.val) {                while(cur.next != null && cur.next.val == cur.val) {                    cur = cur.next;                }                //pre不需后移，因为node.next也可能是重复的                pre.next = cur.next;            }            //当前节点与下一节点            else {                pre.next = cur;                pre = pre.next;            }            cur = cur.next;        }        return root.next;    }}

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