9_leetcode_remove Nth Node from End

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.

//1.链表为空或n不大于0 ； 2.设置两个指针，fast和slow。fast先走n步。3.删除节点考虑三种情况，当倒数第n个节点是头节点， 或者为尾节点，或者时中间节点。三种情况 4.最后一种情况，n大约链表中节点的个数

        ListNode *removeNthFromEnd(ListNode *head, int n)    {        if(head == NULL || n <= 0)            return head;                ListNode* slow = head;        ListNode* fast = head;        int i = 0;        while(i < n)        {            fast = fast->next;            i++;            if(fast == NULL)                break;        }                if(i == n)        {            if(fast == NULL)            {                ListNode* resultHead = head->next;                delete head;                return resultHead;            }            else            {                while(fast->next)                {                    slow = slow->next;                    fast = fast->next;                }                ListNode* deleteNode = slow->next;                slow->next = deleteNode->next;                delete deleteNode;                return head;            }        }        else        {            return head;        }    }