题目: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。
非递归实现:
/*
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 pre = pHead;
ListNode cur = pre.next;
while(cur != null) {
if(cur.val == pre.val) { //pre, cur重复结点
while(cur != null && cur.val == pre.val) {
cur = cur.next;
}
if(cur == null)
return null;
pre = cur;
pHead = pre; //改变头结点
cur = pre.next;
} else {
if(cur.next != null && cur.val == cur.next.val) { //中间重复结点
while(cur.next != null && cur.val == cur.next.val)
cur = cur.next; //去掉重复结点
} else { //不重复结点
pre = cur;
}
cur = cur.next;
}
pre.next = cur;
}
return pHead;
}
}