题目
Given a linked list, remove the nth node from the end of list and return its head.
For example,
Given linked list: 1->2->3->4->5, and n = 2.
After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:
Given n will always be valid.
Try to do this in one pass.
分析
简单的链表操作，不多说，直接上代码。
实现一
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
// if(head==NULL) return NULL;
int length=1;
ListNode * p=head;
while(p->next!=NULL){
p = p->next;
length++;
}
if(length>n){
p=head;
for(int i=0; i<length-n-1; i++){
p = p->next;
}
p->next=p->next->next;
}
else{
head=head->next;
}
return head;
}
};
思考一
提交之后发现自己只打败了2.8%的用户，内心是崩溃的。简单的链表删除还有什么花头吗。遂查看别人的代码以及题解，发现也差不多啊。后来发现是提交时段的问题，重新用这段代码提交一次就好了=_=
