每个重复的元素只留下一个,通过修改next指针的位置来实现
/**
* 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) {
auto dummy = new ListNode(0);
dummy->next = head;
auto left = head;
while(left)
{
auto right = left->next;
while(right && left->val == right->val)
{
right = right->next;
}
left->next = right;
left = left->next;
}
return dummy->next;
}
};
第二次写题
/**
* 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) {
ListNode* dummy = new ListNode(0);
dummy->next = head;
while(head)
{
auto pt = head->next;
while(pt && pt->val == head->val) pt = pt->next;
head->next = pt;
head = head->next;
}
return dummy->next;
}
};