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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.
简单的指针操作。
/** * 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 *pn=head; int sum=1; while(pn->next){ pn=pn->next; sum++; } int tn=sum-n+1; if(tn==1){ return head->next; } pn=head; int i=1; while(i!=tn-1){ pn=pn->next; i++; } pn->next = pn->next->next; return head; } };
leetcode 19. Remove Nth Node From End of List（链表）
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原文地址：http://www.cnblogs.com/zywscq/p/5403054.html
