Insert into a Cyclic Sorted List

最新推荐文章于 2020-09-07 10:48:56 发布

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本文介绍如何在一个已排序的无环链表中插入一个值，并保持链表的有序性。给出的节点可以是链表中的任意节点。文章提供了一段 Java 和 Python 的实现代码。

Given a node from a cyclic linked list which has been sorted, write a function to insert a value into the list such that it remains a cyclic sorted list. The given node can be any single node in the list. Return the inserted new node.

Notice

3->5->1 is a cyclic list, so 3 is next node of 1.
3->5->1 is same with 5->1->3

Have you met this question in a real interview?

Yes

Example

Given a list, and insert a value 4:
3->5->1

Return 5->1->3->4

java

/**
 * Definition for ListNode
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */


public class Solution {
    /*
     * @param node: a list node in the list
     * @param x: An integer
     * @return: the inserted new list node
     */
    public ListNode insert(ListNode node, int x) {
        // write your code here
        ListNode n = new ListNode(x);
        if (node == null) {
            n.next = n;
            return n;
        }
        ListNode curr = node.next;
        ListNode prev = node;
        while (curr != node) {
            if (prev.val <= x && curr.val >= x) {
                break;
            }
            if (prev.val > curr.val && (prev.val < x || curr.val > x)) {
                break;
            }
            curr = curr.next;
            prev = prev.next;
        }
        prev.next = n;
        n.next = curr;
        return node;
    }
}

python

"""
Definition of ListNode
class ListNode(object):

    def __init__(self, val, next=None):
        self.val = val
        self.next = next
"""


class Solution:
    """
    @param: node: a list node in the list
    @param: x: An integer
    @return: the inserted new list node
    """
    def insert(self, node, x):
        # write your code here
        n = ListNode(x)
        if node is None:
            n.next = n
            return n
        prev, curr = node, node.next
        while curr != node:
            if prev.val <= x and curr.val >= x:
                break
            if prev.val > curr.val and (prev.val < x or curr.val > x):
                break
            curr, prev = curr.next, prev.next
        prev.next = n
        n.next = curr
        return node