删除链表的倒数第n个结点 Remove Nth Node From End of List

题目源自于Leetcode。

题目：Given a linked list, remove the nth node from the end of list and return its head.只允许过一遍。

思路：注意当原链表节点个数小于n和等于n的时候这两种特殊情况。

/**
 * 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) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(n<=0)
            return head;
        ListNode *pre, *p;
        pre = p = head;
        int i;
        for(i=0;i<n;i++)
            if(p->next!=NULL)
                p = p->next;
            else
                break;
        if(i<n-1)//原链表总个数小于n个
            return head;
        if(i == n-1)//原链表总个数等于n，即要删除链首结点
        {
            ListNode *tmp1 = head;
            ListNode *tmp2 = head->next;
            delete tmp1;
            return tmp2;
        }
        while(p->next!=NULL)
        {
            p = p->next;
            pre = pre->next;
        }
        ListNode *tmp = pre->next;
        pre->next = tmp->next;
        delete tmp;
        return head;
    }
};