Remove Nth Node From End of List

Remove Nth Node From End of ListJan 28 '12

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.

思路：设一个快指针一个慢指针，快指针提前于慢指针n，然后两者同时向后移动，那么当快指针到达链表尾端时，慢指针刚好到达倒数第n位，去掉这个节点即可。代码如下：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */

     ListNode *removeNthFromEnd(ListNode *head, int n) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        ListNode *fast = head;        ListNode *slow = head;                for(int i = 0;i < n; i++)            fast = fast->next;                   if(fast == NULL)             return head->next;                while(fast->next!=NULL)        {            fast = fast->next;            slow = slow->next;        }                slow->next = slow->next->next;        return head;    }

32 milli secs