[leetcode] Remove Nth Node From End of List-优快云博客

本文介绍了一种使用双指针技术在单次遍历中删除链表倒数第N个节点的方法。通过设置两个相隔N-1距离的指针，当后方指针到达链表尾部时，前方指针正好指向待删除节点的前一个位置，从而实现高效删除。

2019独角兽企业重金招聘Python工程师标准>>>

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

For 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.
Try to do this in one pass.

思路1：双指针思想，两个指针相隔n-1，每次两个指针向后一步，当后面一个指针没有后继了，前面一个指针就是要删除的节点。注意：删除节点时需要删除指针的前驱pre；增加dummy head处理删除头节点的特殊情况。

public class Solution {
	public ListNode removeNthFromEnd(ListNode head, int n) {
		if (head == null)
			return null;
		ListNode dummyHead = new ListNode(-1);
		dummyHead.next = head;
		ListNode s = head;
		ListNode t = head;
		n = n - 1;
		for (int i = 0; i < n; i++) {
			t = t.next;
		}
		ListNode preS = dummyHead;
		while (t.next != null) {
			preS = preS.next;
			s = s.next;
			t = t.next;
		}
		// remove s
		preS.next = s.next;
		s.next = null;
		return dummyHead.next;

	}

	public static void main(String[] args) {
		ListNode head = new ListNode(1);
		head.next = new ListNode(2);
		head.next.next = new ListNode(3);
		head.next.next.next = new ListNode(4);
		head.next.next.next.next = new ListNode(5);
		ListNode newHead = new Solution().removeNthFromEnd(head, 1);

	}



}

转载于:https://my.oschina.net/jdflyfly/blog/283678