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.
思路一： 取到 整个link的长度l，然后移删除第l-n 个节点。
思路二：取两个指针，front和back，front优先于back n 步，当front指向最后一个节点时，back指向从后数n+1节点，然后删除第n个节点。
注意code中使用了dummy节点，避免了如果删除head节点的复杂逻辑判断。也避免了front指向NULL的情况，非常好用，屡试不爽。
class Solution { public: ListNode *removeNthFromEnd(ListNode *head, int n) { ListNode dummy(-1); dummy.next = head; ListNode* front = &dummy; ListNode* back = &dummy; while(n) { front = front->next; n--; } while(front->next) { front = front->next; back = back->next; } back->next = back->next->next; return dummy.next; } };
[LeetCode] Remove Nth Node From End of List
原文：http://www.cnblogs.com/diegodu/p/4277045.html