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  • Leetcode-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.

    分析是:1.逆序链表,时间开销貌似较大,而且也很麻烦,和扫描两次链表没啥区别,不做

               2. 观察,如果放两个指针,p 和q,如果q指向链表尾,p指向第n个节点的话(反向数节点),那么刚好,也就是说保证p和q指针的距离,扫描链表,等q到链表尾了,p也就是指向了想要remove的节点了

      特殊情况就是如果要删的是head节点(原因在于p,q初始都从head开始)

      另外要注意的是删节点需要知道前驱,所以遍历到待删除节点的前驱就不要再扫描了

     1         public ListNode removeNthFromEnd(ListNode head, int n) {
     2              ListNode p = head;
     3              ListNode q = head;
     4              
     5              //Init to make sure the distance between p and q is (1,n)
     6              int counter  = 1;
     7              while(counter < n){
     8                 q  = q.next;
     9                 counter ++;
    10              }
    11              
    12              //if n points to the head
    13              if(q.next == null){
    14                  return head.next;
    15              }
    16              
    17              //remove a node needs to know its previous node
    18              while(q.next.next !=null){
    19                  p = p.next;
    20                  q = q.next;
    21              }
    22              //remove the node 
    23               p.next = p.next.next;
    24              
    25              return head;
    26         
    27         }
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  • 原文地址:https://www.cnblogs.com/dijkstra-c/p/3967223.html
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