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]
提示:
- 链表中结点的数目为
sz
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz
**进阶:**你能尝试使用一趟扫描实现吗?
代码:
objectivec
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* removeNthFromEnd(struct ListNode* head, int n) {
struct ListNode *pTemp=malloc(sizeof(struct ListNode));
pTemp->val=0;
pTemp->next = head;
struct ListNode *pFast=head,*pSlow=pTemp;
while(n!=0){
n--;
pFast=pFast->next;
}
while(pFast){
pFast=pFast->next;
pSlow=pSlow->next;
}
pSlow->next=pSlow->next->next;
struct ListNode *p = pTemp->next;
free(pTemp);
return p;
}
没什么意外,双指针类型的题,掌握规律不要漏掉细节就OK啦