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  • leetcode[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.

    /**
     * Definition for singly-linked list.
     * struct ListNode {
     *     int val;
     *     ListNode *next;
     *     ListNode(int x) : val(x), next(NULL) {}
     * };
     */
    class Solution {
    public:
        ListNode *removeNthFromEnd(ListNode *head, int n) {
            if(head==NULL||head->next==NULL)return NULL;
            ListNode *former=head;
            ListNode *later=head;
            ListNode *temp=NULL;
            int count=0;
            while(former->next)
            {
                count++;
                former=former->next;
                if(count>=n)
                {
                    temp=later;
                    later=later->next;
                }
            }
            if(temp==NULL)head=later->next;
            else temp->next=later->next;
            delete later;
            return head;
        }
    };
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  • 原文地址:https://www.cnblogs.com/Vae1990Silence/p/4283669.html
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