[CrackCode] 2.2 Find the nth to last element of a singly linked list

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最新推荐文章于 2022-01-02 15:21:24 发布

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分类专栏： interview preparation 文章标签： LinkedList algorithm crackcode java

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本文介绍了如何通过设置两个指针，距离为N，来找到单链表中倒数第N个元素。通过移动指针直到右指针到达链表末尾，此时左指针所指向的元素即为所求。

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Implement an algorithm to find the nth to last element of a singly linked list.

==========

Analysis:

Similar problem as "Remove Nth Node From End of List" in LeetCode. Idea is to use two pointers, distance between which is n. So, when the right pointer reached the end of the link list, the left pointer is pointing to the nth to last element.

public class Answer {

	public static LinkedListNode solution(LinkedListNode head, int n){
		if(head == null) return null;
		LinkedListNode left = head;
		LinkedListNode right = head;
		
		for(int i=0; i<n-1; i++){
			if(right.next!=null) right = right.next;
			else return null;
		}
		
		while(right.next!=null){
			left = left.next;
			right = right.next;
		}
		
		return left;
	}
	
	public static void main(String[] args) {
		LinkedListNode head = AssortedMethods.randomLinkedList(10, 0, 10);
		System.out.println(head.printForward());
		int nth = 11;
		LinkedListNode n = solution(head, nth);
		if (n != null) {
			System.out.println(nth + "th to last node is " + n.data);
		} else {
			System.out.println("null");
		}
	}
}

Note: Please download crackcode library before running the program.