给你一个链表,删除链表的倒数第?n
?个结点,并且返回链表的头结点。
示例 1:
输入:head = [1,2,3,4,5], n = 2 输出:[1,2,3,5]
示例 2:
输入:head = [1], n = 1 输出:[]
示例 3:
输入:head = [1,2], n = 1 输出:[1]
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode* temp = head;
stack<ListNode*> st;
while(head!=nullptr)
{
st.push(head);
head=head->next;
}
ListNode* dump=nullptr;
while(n!=0)
{
dump = st.top();
st.pop();
n--;
}
if(dump==temp)
{
dump=dump->next;
return dump;
}
else
{
if(st.size()!=0)
{
dump= st.top();
dump->next=dump->next->next;
}
}
return temp;
}
};