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：
用快慢指针的思想，快指针先移动n个结点，然后快慢指针一起移动，直到快指针到达末尾，然后
快指针的下一个即为要删除的结点。
ListNode* removeNthFromEnd(ListNode* head, int n) { if(head == NULL || n<=0)return head; ListNode* dumb = new ListNode(0); dumb->next = head; ListNode* p = dumb; head = dumb; while(p !=NULL && n--) { p = p->next; } while(p !=NULL && p->next != NULL) { p = p->next; head = head->next; } if(head->next != NULL) head->next = head->next->next; return dumb->next; }
[leetcode-19-Remove Nth Node From End of List]
原文：http://www.cnblogs.com/hellowooorld/p/6682828.html