LeetCode:Remove Nth Node From End of List

Remove Nth Node From End of List

Total Accepted: 82072 Total Submissions: 296476 Difficulty: Easy

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：

1.先用一个指针q向后走n步（如果链表够长）。

2.再让p(=head),q一起向后走，q走到尾为止。

3.边界条件：

1）如果n > 链表长度，应直接返回head。（实际上OJ并没有这样的数据，并且也不是直接返回head，而是删除了头结点）

2）如果n==链表长度，删除头结点。

3）n < 链表长度的情况最好处理。

code:

/** * 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 *p = head, *q = head;        while( q && n)         {            --n;            q=q->next;        }        if(n) return head;        else if(NULL==q)         {            head=head->next;            return head;        }        while(q->next) {            q=q->next;            p=p->next;        }        ListNode *t = p->next;        p->next = t->next;        free(t);        return head;    }};