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  • leetCode(9):Remove Nth Node From End of List 分类: leetCode 2015-06-18 08:15 109人阅读 评论(0) 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.


    /**
     * 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)
        		return NULL;
        	ListNode* fast=head;
        	ListNode* slow=head;
        	int i=0;
        	while(i<n+1 && fast)
        	{
        		i++;
        		fast=fast->next;
        	}
        	if(!fast && i==n)
        	{
        		ListNode* toBeDeleted=head;
        		ListNode* newHead=toBeDeleted->next;
        		delete toBeDeleted;
        		toBeDeleted=NULL;
        		return newHead;
        	}
        	if(!fast && i<n+1)
        		return head;
        	while(fast)
        	{
        		slow=slow->next;
        		fast=fast->next;
        	}
        	ListNode* toBeDeleted=slow->next;
        	slow->next=toBeDeleted->next;
        	delete toBeDeleted;
        	toBeDeleted=NULL;
        	return head;	
        }
    };


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  • 原文地址:https://www.cnblogs.com/zclzqbx/p/4687108.html
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