【LeetCode】26.Linked List —Remove Nth Node From End of List 从列表末尾删除第n个节点

Given a linked list, remove the n-th node from the end of list and return its head.

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.

Follow up:

Could you do this in one pass?

首先遍历链表，得到链表长度。定义两个指针，一前一后，一同遍历链表，直到先出发的指针定位到目标结点，然后进行删除操作。需要对删除头节点的情况进行特殊处理。

/**
 * 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) {
        ListNode*p=head;
        ListNode*q=head;
        int counter=0;
        while(p!=NULL)
        {
            p=p->next;
            counter++;
        }
        p=head;
        if(counter-n==0)
        {
            head=head->next;
            delete p;
            return head;
        }
        for(int i=0;p!=NULL&&i<counter-n;i++)
        {
            p=p->next;
            if(i>0) q=q->next;
        }
        if(p!=NULL)
        {
            q->next=p->next;
            delete p;
        }
        return head;
    }
};