删除链表的倒数第N个节点
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]
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* removeNthFromEnd(struct ListNode* head, int n){
struct ListNode *fast = head ,*slow = head ;
int i = 0;
for( ; i < n; i++)
fast = fast->next; // 快指针移动n步
while(fast && fast ->next) // 判断快指针 和 快指针的 next 是否为空
{
fast = fast->next;
slow = slow->next;
}
if(fast == NULL) // 如果快指针为空 则删除的是头结点
head = head->next;
else // 慢指针为该节点的前继结点 指向下下个结点
slow->next = slow->next->next;
return head;
}
}