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：双指针，相差n
/** * 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* pFirst = head, *pSecond = head; for(int i = 0; i < n; i++){ pFirst = pFirst->next; } if(pFirst == NULL){ return head->next; } while(pFirst->next){ pFirst = pFirst->next; pSecond = pSecond->next; } pSecond->next = pSecond->next->next; return head; } };
19.Remove Nth Node From End of List(List; Two-Pointers)
原文：http://www.cnblogs.com/qionglouyuyu/p/4687270.html