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
解决方案
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};
*/
class Solution {
public:
ListNode* deleteDuplication(ListNode* pHead)
{
if(!pHead || !pHead->next) return pHead;
ListNode* start = new ListNode(0);
start->next = pHead;
ListNode* pre = start;
while(pre->next)
{
ListNode* cur = pre->next;
while(cur->next && cur->next->val == cur->val)
{
cur = cur->next;
}
if(cur != pre->next)
{
pre->next = cur->next;
}else{
pre = pre->next;
}
}
return start->next;
}
};