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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.代码如下：
/** * 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 == NULL) return head; ListNode *p=head, *q = head; for (int i =0; i next; }else return head; } if (p == NULL) { head = head->next; return head; } while(p->next != NULL){ p = p->next; q = q->next; } q->next = q->next->next; return head; }}; 转载地址：http://ibxti.baihongyu.com/