LeetCode 19. 删除链表的倒数第 N 个结点

题目

19. 删除链表的倒数第 N 个结点

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]

提示：

• 链表中结点的数目为 sz
• 1 <= sz <= 30
• 0 <= Node.val <= 100
• 1 <= n <= sz

思路

算法：
双指针算法 O(n)

1、创建虚拟结点dummy，dummy指向head.next
2、指针first，second初始化均指向dummy，first指针走n步，first,second指针同时向后走，直到first走到末尾时终止
3、这样找到了删除结点的前一个结点，让前一个结点指向删除结点的后一个结点即可

时间复杂度：只遍历一次链表，所以总的时间复杂度为 O(n) .

代码

C++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* removeNthFromEnd(ListNode* head, int n) {
        ListNode* dummy = new ListNode(-1);
        dummy->next = head;
        ListNode* first = dummy;
        ListNode* second = dummy;
        for(int i = 0; i <= n; i++) first = first->next;
        while(first){
            first = first->next;
            second = second->next;
        }
        second->next = second->next->next;
        return dummy->next;
    }
};

python3代码：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        # Two pointer, One pass
        dummy = ListNode(0)
        dummy.next = head
        fast = slow = dummy
        for _ in range(n):
            fast = fast.next
        while fast and fast.next:
            fast = fast.next
            slow = slow.next
        slow.next = slow.next.next
        return dummy.next