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）为了避免反复计数，採用双指针，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。经常使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;
    }
};