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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.
常规的解法，就是先轮训一次，计算其list的总长度len，再计算出需要要删除的index（len - n），但这需要两次循环。代码如下：
class Solution { public: ListNode* removeNthFromEnd(ListNode* head, int n) { int cnt = 0; ListNode* node = head; while (node) { ++cnt; node = node->next; } int index = cnt - n; if (0 == index) { head = head->next; } else { node = head; while (--index) { node = node->next; } node->next = node->next->next; } return head; } };
上面的解法中，要先计算出len，才能算出index。我们可以用另外一种方法来计算出index，先循环n次，让node自增。再新建一个指针slow指向head，之后再接着node和slow自增，node为空时，slow也就刚好到了要删除的点了。代码如下：
ListNode* removeNthFromEnd(ListNode* head, int n) { ListNode **slow = &head, *fast = head; while (--n) fast = fast->next; while(fast->next) { slow = &((*slow)->next); fast = fast->next; } *slow = (*slow)->next; return head; }
19 Remove Nth Node From End of List
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原文地址：http://www.cnblogs.com/styshoo/p/4657860.html
