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
非递归的代码思路:
- 首先添加一个头节点,以方便碰到第一个,第二个节点就相同的情况
2.设置 pre ,last 指针, pre指针指向当前确定不重复的那个节点,而last指针相当于工作指针,一直往后面搜索。
参考大神的思路而来的代码。
python
# -*- coding:utf-8 -*-
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def deleteDuplication(self, pHead):
# write code here
if (pHead == None or pHead.next == None):
return pHead
Head = ListNode(0)
Head.next = pHead
pre = Head
last = Head.next
while(last != None):
if(last.next!=None and last.val == last.next.val):
while(last.next!=None and 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
c++
class Solution {
public:
ListNode* deleteDuplication(ListNode* pHead)
{
//if(pHead==NULL)
// return NULL;
ListNode* pPreNode = NULL;
ListNode* pNode = pHead;
while(pNode!=NULL)
{
ListNode* pNext = pNode->next;
bool needDelete = false;
if(pNext!=NULL&&pNext->val==pNode->val)
needDelete = true;
if(!needDelete)
{
pPreNode = pNode;
pNode = pNode->next;
}
else
{
while(pNext!=NULL&&pNext->val==pNode->val)
{
pNext = pNext->next;
}
if(pPreNode==NULL)
pHead = pNext;
else
pPreNode->next = pNext;
pNode = pNext;
}
}
return pHead;
}
};