Leetcode [linked list]: Remove nth node from end of list

Given a linked list, remove the nth node from the end of list and return its head.
For example,
Given linked list: 1->2->3->4->5, and n = 2.
After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:
Given n will always be valid.
Try to do this in one pass.

I have two solutions for this question. Time ComplexityO(N), Space complexity is different, one isO(N).O(1 ).

In addition, I think it is important to delete the deleted node to prevent memory leakage.

The space complexity is O( N).

Use a vector to store the addresses of all nodes. The Code is as follows:

C++ code    ListNode *removeNthFromEnd(ListNode *head, int n) {        vector<ListNode *> list;        for (ListNode *iter = head; iter != NULL; iter = iter->next)            list.push_back (iter);        if (n == list.size()) {            head = list[0]->next;            delete list[0];            return head;        }        list[list.size() - n - 1]->next = list[list.size() - n]->next;        delete list[list.size() - n];        return list[0];    }

The space complexity is O(1) Solution

Set two pointers, fast faster than N nodes. The Code is as follows:

C++ code    ListNode *removeNthFromEnd(ListNode *head, int n) {        ListNode *fast = head, *low = head, *delNode;        for (int i = 0; i < n; ++i)            fast = fast->next;        if (!fast) {            delNode = head;            head = head->next;            delete delNode;            return head;        }        while (fast->next) {            fast = fast->next;            low  = low ->next;        }        delNode = low->next;        low->next = low->next->next;        delete delNode;        return head;    }

Leetcode [linked list]: Remove nth node from end of list