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.
考虑用双链表，先让第一个链表走n-1步，然后让第二个链表从头开始两个链表一起走，直到第一个走到头，此时第二个链表刚好走到要删除的节点，保存删除节点的上一个节点即可，但是要考虑边界条件，这里考虑了n输入不合法的问题
/** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ struct ListNode *removeNthFromEnd(struct ListNode *head, int n) { if(head == NULL || n<= 0) return NULL; struct ListNode *pre = NULL; struct ListNode *p = head; struct ListNode *q = head; n--; while(p->next != NULL && n>0){ p = p->next; n--; } if(n>0) return NULL; while(p->next != NULL){ pre = q; p = p->next; q = q->next; } if(pre == NULL){ head = q->next; }else{ pre->next = q->next; } return head; }
看到还有一种方法是走到要删除的节点时，将它的下一个节点值复制到当前节点，然后删除它的下一个节点，这种方法也可以。但是要判断倒数第一个节点的问题。
Remove Nth Node From End of List
原文：http://www.cnblogs.com/zhhc/p/4370084.html