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  • [Leetcode] remove nth node from the end of list 删除链表倒数第n各节点

    Given a linked list, remove the n th 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.

    这题比较简单,使用快慢指针,找到倒数第n个结点的前驱即可,然后连接其前驱和后继即可,值得注意的是,两个while的条件之间的关系。代码如下:

     1 /**
     2  * Definition for singly-linked list.
     3  * struct ListNode {
     4  *     int val;
     5  *     ListNode *next;
     6  *     ListNode(int x) : val(x), next(NULL) {}
     7  * };
     8  */
     9 class Solution {
    10 public:
    11     ListNode *removeNthFromEnd(ListNode *head, int n) 
    12     {
    13         ListNode *nList=new ListNode(-1);
    14         nList->next=head;
    15         ListNode *pre=nList;
    16         ListNode *fast=head;
    17         ListNode *slow=head;
    18         int num=0;
    19 
    20         while(num++<n)
    21         {
    22             fast=fast->next;
    23         }   
    24         while(fast->next)
    25         {
    26             pre=pre->next;
    27             slow=slow->next;
    28             fast=fast->next;
    29         } 
    30         pre->next=slow->next;
    31 
    32         return nList->next;
    33     }
    34 };
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  • 原文地址:https://www.cnblogs.com/love-yh/p/7050488.html
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