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pq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;
    }
}

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