Leetcode 题目解析之 Remove Nth Node From End of List

原创

ruochen

发布于 2022-01-14 11:42:05

1.3K0

发布于 2022-01-14 11:42:05

文章被收录于专栏：若尘的技术专栏

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.

题目中说n是合法的，就不用对n进行检查了。用标尺的思想，两个指针相距为n-1，后一个到表尾，则前一个到n了。(p为second，q为first)

指针p、q指向链表头部；
移动q，使p和q差n-1；
同时移动p和q，使q到表尾；
删除p。

public ListNode removeNthFromEnd(ListNode head, int n) {

    if (head == null || head.next == null) {
        return null;
    }

    ListNode first = head;
    ListNode second = head;

    for (int i = 0; i < n; i++) {
        first = first.next;
        if (first == null) {
            return head.next;
        }
    }

    while (first.next != null) {
        first = first.next;
        second = second.next;
    }

    second.next = second.next.next;

    return head;
}

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node.js