19. Remove Nth Node From End of List

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

Follow up: Could you do this in one pass?

• new solution update 2022.02.13

c++

/**
 * 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) {
        ListNode* tail = head;
        for(int i = 0; i < n; i++) {
            tail = tail->next;
        }
        ListNode* dummy = new ListNode(-1,head);
        ListNode* prev = dummy;
        while(tail) {
            prev = prev->next;
            tail = tail->next;
        }
        prev -> next = prev->next->next;
        return dummy->next;
    }
};

c++

/**
 * 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) {
        ListNode* tail = head;
        ListNode* target = head;
        while(n--) {
            tail = tail->next;
        }
        if(tail == nullptr) {
            return head -> next;
        }
        
        while(tail->next != nullptr) {
            target = target->next;
            tail = tail->next;
        }
        target->next = target->next->next;
        
        return head;
    }
};