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.

 

 to see which companies asked this question

我使用两个指针，两个指针之间的距离保持相差n

较慢的指针指向的就是需要删除的结点

ListNode* removeNthFromEnd(ListNode* head, int n) {    ListNode dummy(0);    dummy.next = head;    ListNode *slow = &dummy, *fast = &dummy;    while (n--)        fast = fast->next;    while (fast->next != nullptr)    {        slow = slow->next;        fast = fast->next;    }    ListNode *temp = slow->next;    slow->next = slow->next->next;    delete temp;    return dummy.next;}