Remove Nth Node From End of List -LeetCode

最新推荐文章于 2021-07-11 21:56:15 发布

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本文介绍如何在单链表中找到并删除倒数第N个节点，通过迭代实现一次遍历完成操作。

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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.

/**
 * 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)
		    return NULL;
	    ListNode *fir = head;
	    ListNode *sec = head;
    	int i = 0;
    	for (i = 0; i < n + 1 && fir != NULL; i++)
	    	fir = fir->next;
    	if (i < n + 1)
    	{
    		head = head->next;
    		return head;
    	}
    	while(fir!=NULL)
    	{
    		fir = fir->next;
    		sec = sec->next;
    	}
	
    	sec->next = sec->next->next;
    	return head;
    }
};