# 19. Remove Nth Node From End of List

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

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.

Follow up:

Could you do this in one pass?

# Solution

Approach 1: Two pass.

Approach 2: One pass.

# Code (Python)

Approach 1:

Approach 2:

# Code (C++)

Approach 1:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
// Two pass.
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode dummyHead(0);
        dummyHead.next = head;
        ListNode *node = head;
        int len = 0;
        while (node)
        {
            len++;
            node = node->next;
        }
        len -= n;
        node = &dummyHead;
        for (int i = 0; i < len; ++i)
        {
            node = node->next;
        }
        node->next = node->next->next;
        return dummyHead.next;
    }
};

Approach 2:

// One pass.
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode dummyHead(0);
        dummyHead.next = head;
        ListNode *slow = &dummyHead;
        ListNode *fast = &dummyHead;
        for (int i = 0; i < n; ++i)
            fast = fast->next;
        while (fast && fast->next)
        {
            slow = slow->next;
            fast = fast->next;
        }
        slow->next = slow->next->next;
        return dummyHead.next;
    }
};