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
【思路】递归
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};
*/
class Solution {
public:
ListNode* deleteDuplication(ListNode* pHead)
{
if((pHead == NULL)||(pHead != NULL && pHead->next == NULL))
return pHead;
ListNode* current;
if(pHead->val == pHead->next->val){
current = pHead->next->next;
while(current != NULL && current->val == pHead->val)
current = current->next;
return deleteDuplication(current);
}else{
current = pHead->next;
pHead->next = deleteDuplication(current);
return pHead;
}
}
};