个人向
前后双指针
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?
来源:力扣(LeetCode) 链接:https://leetcode.cn/problems/remove-nth-node-from-end-of-list 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode* dummy = new ListNode(0, head);
ListNode* front = dummy, * back = dummy;
for (int i = 1; i <= n + 1; i++)
back = back->next;
while (back) {
front = front->next;
back = back->next;
}
front->next = front->next->next;
return dummy->next;
}
};
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