题目
给你一个链表,删除链表的倒数第 n 个结点,并且返回链表的头结点。
示例 1:
输入:head = [1,2,3,4,5], n = 2
输出:[1,2,3,5]
示例 2:
输入:head = [1], n = 1
输出:[]
示例 3:
输入:head = [1,2], n = 1
输出:[1]
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/remove-nth-node-from-end-of-list
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提示:
链表中结点的数目为 sz
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz
题解思路
代码
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* middleNode(ListNode* head) {
ListNode* middle_node = head;
int count=1;
int current_index = 1;
while(head)
{
count = count+1;
head = head->next;
}
int cc = 1;
while(middle_node)
{
if(count%2==0 && cc==count/2)
{
return middle_node;
break;
}
else if(count%2==1 && cc==count/2)
{
return middle_node->next;
break;
}
cc+=1;
middle_node = middle_node->next;
}
return middle_node;
}
};