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  • [LeetCode19]Remove Nth Node From End of List

    题目:

    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.

    给定一个单链表,移除倒数第n个Node,返回链表的head

    思路:维护两个指针slow,fast,fast先移动n步,然后slow、fast同时移动,直到fast为空为止,删除slow后面的结点即可

    代码:

     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         ListNode *preHead = new ListNode(0);
    13         ListNode *fast = preHead, *slow = preHead;
    14         slow->next = head;
    15         for(int i = 0; i <= n; ++i)
    16             fast = fast->next;
    17         while(fast)
    18         {
    19             fast = fast->next;
    20             slow = slow->next;
    21         }
    22         slow->next = slow->next->next;
    23         return preHead->next;
    24     }
    25 };
    
    
    
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  • 原文地址:https://www.cnblogs.com/zhangbaochong/p/5186060.html
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