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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.
题意：移除链表的倒数第n个元素
思路：
两个指针p, q，
p先走n步，然后p，q一起走，当p走到尾的时候，q->next就是要删除的节点
复杂度： 时间O(n)，空间O(1)
/**
* 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 dummy(0);
dummy.next = head;
ListNode *p, *q;
p = q = &dummy;
while(n--) p= p->next;
while(p->next){
p = p->next;
q = q->next;
}
ListNode *to_delete = q->next;
q->next = to_delete->next;
delete to_delete;
return dummy.next;
}
};Leetcode 线性表 Remove Nth Node From End of List,布布扣,bubuko.com
Leetcode 线性表 Remove Nth Node From End of List
原文：http://blog.csdn.net/zhengsenlie/article/details/25797127