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。

进阶：你能尝试使用一趟扫描实现吗？

  • 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

Solution

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

/**
 * 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) {
        if (n < 1) {
            return head;
        }
        ListNode *fast = head;
        while (fast != nullptr && n > 0) {
            fast = fast->next;
            n--;
        }
        if (n != 0 && fast == nullptr) {
            return head;
        }

        if (n == 0 && fast == nullptr) {
            return head->next;
        }

        ListNode *last = head;
        while (fast->next != nullptr) {
            fast = fast->next;
            last = last->next;
        }
        ListNode *temp = last->next;
        last->next = temp->next;
        temp->next = nullptr;
        delete temp;
        return head;
    }
};

int main() {
    ListNode *list = new ListNode(0);

    ListNode *temp = list;
    for (int i = 1; i < 6; i++) {
        temp->next = new ListNode(i);
        temp = temp->next;
    }
    list = list->next;
    temp = Solution().removeNthFromEnd(list, 6);
    while (temp != nullptr) {
        std::cout << temp->val << " ";
        temp = temp->next;
    }
    std::cout << std::endl; 
}