19. Remove Nth Node From End of List

Problem:

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:

Set a pointer p1 for iterating, and p2 which is n nodes behind, pointing at the (n+1)-th node from the end of list.

Boundaries that should be awared of:

• p2 could be one node before head, which means the head should be removed.

• p2 could be larger than the length of the list (Though the description says n will always be valid, we take care of it anyway).

• It should be p1.next touches the end rather than p1 because we want p1 pointing at the last node.

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} n
 * @return {ListNode}
 */
let removeNthFromEnd = function (head, n) {
  let p1 = head;
  while (p1 && n--) {
    p1 = p1.next;
  }

  if (!p1) {
    return n ? head : head.next;
  }

  let p2 = head;
  while (p1.next) {
    p1 = p1.next;
    p2 = p2.next;
  }

  p2.next = p2.next.next;

  return head;
};

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☆*: .｡. o(≧▽≦)o .｡.:☆☆: .｡. o(≧▽≦)o .｡.:☆☆: .｡. o(≧▽≦)o .｡.:*☆

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☆*: .｡. o(≧▽≦)o .｡.:☆☆: .｡. o(≧▽≦)o .｡.:*☆

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