[数据结构与算法]力扣 删除链表的倒数第N个结点

Given the head of a linked list, remove the nth node from the end of the list and return its head.

Example 1:

Input: head = [1,2,3,4,5], n = 2
Output: [1,2,3,5]

Example 2:

Input: head = [1], n = 1
Output: []

Example 3:

Input: head = [1,2], n = 1
Output: [1]

Constraints:

The number of nodes in the list is sz.
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz

Follow up: Could you do this in one pass?

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) {
        int len = 1;
        ListNode* h1 = head;
        while(h1->next){
            h1 = h1->next;
            len++;/* the length of List*/
        }
        if(n == len){
            
            return head->next;
        }
        int x = len-n;
        h1 = head;
        while(x!=1){
            x--;
            h1 = h1->next;
        }
        h1->next = h1->next->next;
        return head;
    }
};