[LeetCode] 2487. Remove Nodes From Linked List

You are given the head of a linked list.

Remove every node which has a node with a greater value anywhere to the right side of it.

Return the head of the modified linked list.

Example 1:

Input: head = [5,2,13,3,8]
Output: [13,8]
Explanation: The nodes that should be removed are 5, 2 and 3.

• Node 13 is to the right of node 5.
• Node 13 is to the right of node 2.
• Node 8 is to the right of node 3.

Example 2:
Input: head = [1,1,1,1]
Output: [1,1,1,1]
Explanation: Every node has value 1, so no nodes are removed.

Constraints:
The number of the nodes in the given list is in the range [1, 105].
1 <= Node.val <= 105

从链表中移除节点。

给你一个链表的头节点 head 。 移除每个右侧有一个更大数值的节点。 返回修改后链表的头节点 head 。

思路

根据题意，如果某个 node 的右侧有一个比他 val 更大的 node，需要把这个 node 删除。那么这里我们可以反过来思考，如果我们从右往左遍历整个链表，我们可以先把第一个节点的 val 当做最大值，记为 max，再往左遍历的时候，如果当前节点值比 max 小，则把当前节点移除；否则把当前节点的节点值记为 max，继续往左遍历。这样做的好处是，我们只需要遍历一次链表，就可以把所有需要删除的节点都删除掉。不过我们需要将 input 链表整个反转一次，遍历一次，再反转回去。

复杂度

时间O(n)
空间O(1)

代码

Java实现

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeNodes(ListNode head) {
        // corner case
        if (head == null || head.next == null) {
            return head;
        }

        // normal case
        head = reverse(head);
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode cur = dummy;
        int max = 0;
        while (cur.next != null) {
            if (cur.next.val < max) {
                cur.next = cur.next.next;
            } else {
                max = cur.next.val;
                cur = cur.next;
            }
        }
        head = reverse(head);
        return head;
    }

    private ListNode reverse(ListNode head) {
        ListNode cur = head;
        ListNode pre = null;
        while (cur != null) {
            ListNode next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
        }
        return pre;
    }
}