LeetCode:Remove Nth Node From End Of First(java)

最新推荐文章于 2022-10-29 13:05:33 发布

原创最新推荐文章于 2022-10-29 13:05:33 发布 · 149 阅读

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本文介绍了一个LeetCode上的经典链表操作问题——删除链表中倒数第N个节点。通过双指针技巧，实现了仅遍历一次链表即可完成删除操作的方法。文章提供了完整的Java代码实现，并通过实例演示了算法的有效性。

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package LeetCode_LinkedList;



/**题目：
 *      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?
 */

public class RemoveNthFromEnd {
    public ListNode RemoveNthFromEnd(ListNode head,int n){
        ListNode result = new ListNode(0);
        result.next = head;
        ListNode first = result;
        ListNode second = result;
        for (int i = 1; i <= n+1; i++) {
            first = first.next;
        }
        while (first != null) {
            first = first.next;
            second = second.next;
        }
        second.next = second.next.next;
        return result.next;
    }

    public static void main(String[] args) {
        ListNode node1 = new ListNode(1);
        ListNode node2 = new ListNode(2);
        ListNode node3 = new ListNode(3);
        ListNode node4 = new ListNode(4);
        ListNode node5 = new ListNode(5);

        node1.next = node2;
        node2.next = node3;
        node3.next = node4;
        node4.next = node5;

        RemoveNthFromEnd test = new RemoveNthFromEnd();
        ListNode result = test.RemoveNthFromEnd(node1,2);
        while (result != null) {
            System.out.print(result.val + " ");
            result = result.next;
        }
    }
}