leetcode 题解 || Remove Nth Node From End of List 问题

problem：

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.

删除单链表的倒数第n个节点

thinking：

（1）这里的 head 是头指针，指向第一个结点！！！别搞混了。

（2）为了避免重复计数，采用双指针，先让第一个指针走n-1步，再一起走，这样，等前面指针走到最后一个非空结点时，后面一个指针正好指向待删除结点的前驱！！！

（3）延伸：

头结点不是必须的，一般不用，常用的是用一个头指针head指向第一个元素结点！！！！！这道题就是！！！！！

/** * 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) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if (head == NULL)            return NULL;        ListNode *pPre = NULL;        ListNode *p = head;        ListNode *q = head;        for(int i = 0; i < n - 1; i++)            q = q->next;        while(q->next)        {            pPre = p;            p = p->next;            q = q->next;        }        if (pPre == NULL)        {            head = p->next;            delete p;        }        else        {            pPre->next = pPre->next->next;            delete p;        }        return head;    }};