Remove Nth Node From End of List

最新推荐文章于 2024-06-20 12:29:29 发布

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最新推荐文章于 2024-06-20 12:29:29 发布

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本文介绍了一种高效算法，用于从链表中删除倒数第N个节点，仅通过一次遍历实现。提供了完整的代码示例，并解释了如何处理不同情况，如删除头节点等。

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Given a linked list, remove the nth 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.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        if (head == null || n < 1) {
        	return head;
        }
        ListNode cur = head;
        while (cur != null) {
        	n--;
        	cur = cur.next;
        }
        if (n == 0) {
        	head = head.next;
        }
        while (n < 0) {
        	cur = head;
        	while (++n != 0) {
        		cur = cur.next;
        	}
        	cur.next = cur.next.next;
        }
        return head;
    }
}