Description
Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
Example
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.
思路
- 链表去重,一旦重复,所有的重复节点都删除
- 注意删除后为空的情况,和头结点删除的时候
代码
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* deleteDuplicates(ListNode* head) {
if(!head) return NULL;
ListNode *ptr = head, *pNext = head->next;
ListNode *res = new ListNode(0);
ListNode *h = res;
while(pNext){
if(pNext->val == ptr->val){
do{
ListNode *tmp = pNext;
pNext = pNext->next;
delete tmp;
}while(pNext && pNext->val == ptr->val);
delete ptr;
ptr = pNext;
if(ptr)
pNext = ptr->next;
}
else{
res->next = ptr;
res = res->next;
ptr = pNext;
pNext = ptr->next;
}
}
if(ptr && res && res->val != ptr->val){
res->next = ptr;
res = res->next;
}
if(res)
res->next = NULL;
return h->next;
}
};