Remove Nth Node From End of List

2012-11-06 09:26 4人阅读 评论(0) 收藏 编辑 删除

1. 首先我要在纸上，非常非常聪明且迅速且机灵，
2. 给出几个用例，找出边界用例和特殊用例，确定特判条件；在编码前考虑到所有的条件
3. 向面试官提问：问题规模，特殊用例
4. 给出函数头
5. 暴力解，简述，优化。
6. 给出能够想到的最优价
7. 伪代码，同时结合用例
8. 真实代码

Remove Nth Node From End of List

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) {        ListNode* f = head;                for(int i=1; i < n; i++)        {            f = f->next;        }                ListNode* b = head;        ListNode* p;                while(f->next!=NULL)        {            f=f->next;            p=b;            b=b->next;        }                if(b==head)         {            return b->next;        }        else        {             p->next=b->next;             return head;        }    }};