[leetcode]19 删除链表的倒数第 N 个结点 | 链表模拟

​​题目链接​​：https://leetcode.cn/problems/remove-nth-node-from-end-of-list/

该链表中，head节点对应有值

要想知道倒数第几个节点对应正序哪个节点，需要先进行遍历知道整个链表的长度

倒数第n个元素就是第len - n + 1个元素

然后让前一个结点指向当前节点的下一个节点便完成任务

code:

/**
 * 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) {
        int cnt = 0;
        ListNode* h2 = head;
        while(head != nullptr) {
            head = head->next;
            cnt ++;
        }
        // std::cout << cnt << std::endl;
        int pos = cnt - n + 1;
        if(pos == 1) return h2->next;
        ListNode* pre = nullptr;
        head = h2;

        for(int i = 1;i <= cnt;i ++) {
            if(i == pos) {
                /**
                // 1
                if(head->next == nullptr) pre->next = nullptr;
                else {
                    pre->next = head->next;
                }
                **/
                // 2
                pre->next = head->next;
            }
            pre = head;
            head = head->next;
        }
        head = h2;
        return head;
    }
};