剑指offer面试题57 删除链表中重复的结点

解题思路：

设置三个指针，分别是pre,current,later,pre初始值设为null,current指向头结点，later始终指向current的下一个结点。当current指向结点元素与later指向元素不等时，current后移；当current指向结点元素与later指向元素相等时，则一直后移current与later,直到current与later指向元素不等为止。此时更新指针指向，即删除链表中重复节点。

public class Solution {public ListNode deleteDuplication(ListNode pHead) {if (pHead == null) {System.out.println("链表不存在");return null;}//ListNode pre = null;ListNode current = pHead;while (current != null) {ListNode later = current.next;if (later != null) {if (current .val != later.val) {pre = current;current = current.next;} else {while (current != null && later != null && current.val == later.val ) {current = current.next;later = later.next;}//此时循环结束时，current指向重复数字的最后一个，later指向第一个数字//断开指针的连接if (pre != null) {pre.next = later;current = later;} else {pHead = later;current = later;}}} else {//later == null时，current = current.next;}}return pHead;}}