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题目：
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：9ms过集合
/** * 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 ) return head; ListNode dummy(-1); dummy.next = head; ListNode *p1 = &dummy, *p2 = &dummy; for (size_t i = 0; i < n; ++i, p2=p2->next); for (; p2->next; p1=p1->next, p2=p2->next); ListNode *tmp = p1->next; p1->next = p1->next==NULL ? NULL : p1->next->next; delete tmp; return dummy.next; } };
Tips:
双指针技巧：
a. p2先走n步
b. p1和p2一起走，直到p2走到最后一个元素
c. 删除元素
注意再删除元素的时候，保护一下p1->next指针不为NULL。
【Remove Nth Node From End of List】cpp
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原文地址：http://www.cnblogs.com/xbf9xbf/p/4468008.html
