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19. Remove Nth Node From End of List

时间:2016-01-26 21:41:40      阅读:130      评论:0      收藏:0      [点我收藏+]
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

  

19. Remove Nth Node From End of List

原文:http://www.cnblogs.com/kydnn/p/5161567.html

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