LeetCode---(19)Remove Nth Node From End of List

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.

分析
设两个指针p, q，让q 先走n 步，然后p 和 q 一起走，直到q 走到尾节点，删除p->next即可。

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