leetcode---remove-nth-node-from-end-of-list---链表

Given a linked list, remove the n th 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.

/** * 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)     {        if(!head)            return NULL;        ListNode *cur = head;        int cnt = 0;        while(cur)        {            cur = cur->next;            cnt++;        }        if(cnt == 1)            return NULL;        int m = cnt - n;        if(m == 0) // 如果删除的是头部        {            return head->next;        }        cur = head;        cnt = 0;        while(cnt < m-1)        {            cur = cur->next;            cnt++;        }        ListNode *p = cur->next;        if(p)            cur->next = p->next;        else   // 如果删除的是尾部            cur->next = NULL;        delete p;        return head;    }};