19. Remove Nth Node From End of List

https://leetcode.com/problems/remove-nth-node-from-end-of-list/description/

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?

The main idea is find the n-th node from head, then let it and head move together. Finally, when it reaches the end, the head will reach the n-th node from the end.

# Definition for singly-linked list.
class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution(object):
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        target = head
        curr = head.next
        for i in range(n - 1):
            curr = curr.next
        if curr == None:
            return head.next
        while curr.next != None:
            curr = curr.next
            target = target.next
        target.next = target.next.next
        return head