Remove Nth Node From End of List

【题目描述】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.【思路】这道题是参考别人思路的，很长时间没接触链表，这方面比较欠缺。这道题主要是采取双指针的思想，两个指针距离n-1，同时移动两个指针，当后面那个指针移到最后一个的时候，就意味着前面的指针到达了要删除的位置。【代码】

/** * 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* h1=head;        ListNode* h2=head;        ListNode* pre=NULL;        for(int i = 0; i < n - 1; i++)            h2 = h2->next;        while(h2->next!=NULL){            pre=h1;            h1=h1->next;            h2=h2->next;                    }        if(pre==NULL){            head=h1->next;            delete h1;        }        else{            pre->next=h1->next;            delete h1;        }        return head;    }};