LeetCode-in-Cpp

19. Remove Nth Node From End of List

Medium

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?

Solution

/**
 * 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 {
private:
    int n;

    void removeNth(ListNode* node) {
        if (node->next == nullptr) {
            return;
        }
        removeNth(node->next);
        n--;
        if (n == 0) {
            node->next = node->next->next;
        }
    }

public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        this->n = n;
        ListNode* node = new ListNode(0, head);
        removeNth(node);
        ListNode* newHead = node->next;
        delete node; // clean up the temporary node
        return newHead;
    }
};

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