題目

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

簡單的鏈表操作，不多說，直接上代碼。

實(shí)現(xiàn)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

提交之后發(fā)現(xiàn)自己只打敗了2.8%的用戶，內(nèi)心是崩潰的。簡單的鏈表刪除還有什么花頭嗎。遂查看別人的代碼以及題解，發(fā)現(xiàn)也差不多啊。后來發(fā)現(xiàn)是提交時段的問題，重新用這段代碼提交一次就好了=_=