算法1
(暴力枚举) $O(n^2)$
先暴力求整个的长度,然后再去删除
C++ 代码
/**
* 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* dummy = new ListNode(0) ;
dummy->next = head ;
int len = 0 ;
ListNode* cur = dummy;
while(cur->next != nullptr){
len ++ ;
cur = cur->next ;
}
int t = len - n ;
ListNode* tmp = dummy ;
while(t--){
tmp = tmp -> next ;
}
tmp -> next = tmp->next->next ;
return dummy->next ;
}
};
算法2
(暴力枚举) $O(n^2)$
使用快慢指针,详细请看代码
C++ 代码
/**
* 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* dmy = new ListNode(0,head) ;
ListNode* l = head ;
ListNode* r = dmy ;
for(int i=0;i<n;i++){ // 快的先跑n个
l = l -> next ;
}
while(l){//快的跑len-n,慢的也就跑到n-len个前面的一个
l = l->next ;
r = r->next ;
}
r->next = r->next->next;
ListNode* ans = dmy->next ;
delete dmy;
return ans ;
}
};