Remove Nth Node From End of List

描述

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.

  分析

设两个指针p,q，让q先走n步，然后p和q一起走，直到q走到尾节点，删除p->next即可。

代码

// Remove Nth Node From End of List
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode p = dummy, q = dummy;
        for (int i = 0; i < n; i++)  // q先走n步
            q = q.next;
        while(q.next != null) { // 一起走
            p = p.next;
            q = q.next;
        }
        p.next = p.next.next;
        return dummy.next;
    }
}

原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/java/linear-list/linked-list/remove-nth-node-from-end-of-list.html