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;

}
}