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  • 19. Remove Nth Node From End of List

    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.
    

      

      ListNode fast = head, slow = head;
        if(head == null){          //链表中两个指针的思想灰常重要,一定要掌握!
            return null;
        }
        while(n != 0){       
            fast = fast.next;
            n--;
        }
        if(fast == null ){   element is the head  //注意特殊情况!!好好想想什么时候要考虑这些边界值
            return head.next;
        }
        while(fast.next != null){ 
            slow = slow.next;
            fast = fast.next;
        }
        slow.next = slow.next.next;        //被删的节点不用清除么?为什么=null会有错
    

      

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  • 原文地址:https://www.cnblogs.com/kydnn/p/5161567.html
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