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LeetCode Solution, Medium, 708. Insert into a Sorted Circular Linked List

插入排序循環連結序列

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攻城獅
·Jul 13, 2022·

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708. Insert into a Sorted Circular Linked List

題目敘述

Given a Circular Linked List node, which is sorted in ascending order, write a function to insert a value insertVal into the list such that it remains a sorted circular list. The given node can be a reference to any single node in the list and may not necessarily be the smallest value in the circular list.

If there are multiple suitable places for insertion, you may choose any place to insert the new value. After the insertion, the circular list should remain sorted.

If the list is empty (i.e., the given node is null), you should create a new single circular list and return the reference to that single node. Otherwise, you should return the originally given node.

Example 1:

example_1_before_65p.jpeg

Input: head = [3,4,1], insertVal = 2
Output: [3,4,1,2]
Explanation: In the figure above, there is a sorted circular list of three elements. You are given a reference to the node with value 3, and we need to insert 2 into the list. The new node should be inserted between node 1 and node 3. After the insertion, the list should look like this, and we should still return node 3.

Example 2:

example_1_after_65p.jpeg

Input: head = [], insertVal = 1
Output: [1]
Explanation: The list is empty (given head is null). We create a new single circular list and return the reference to that single node.

Example 3:

Input: head = [1], insertVal = 0
Output: [1,0]

Constraints:

  • The number of nodes in the list is in the range [0, 5 * 10**4].
  • -10**6 <= Node.val, insertVal <= 10**6

題目翻譯

給定一個循環的連接序列,要插入一個新的節點,並且需要保持排序的狀態。

解法解析

這題主要使用了 Two Pointer 的解法,因為需要插入節點,所以就需要記住前節點跟當前節點。因為他是排序的,所以只要比較跟前後節點得值就知道插入到哪裡。唯一比較需要注意的是因為是循環的,所以在連結處時。其值的比較就會不同,這時會是相反的,前節點會比當前節點大。

解法範例

Go

/**
 * Definition for a Node.
 * type Node struct {
 *     Val int
 *     Next *Node
 * }
 */

func insert(aNode *Node, x int) *Node {
    if aNode == nil {
        newNode := Node{x, nil}
        newNode.Next = &newNode
        return &newNode
    }

    var prev, curr *Node = aNode, aNode.Next
    for curr != aNode {
        if prev.Val <= x && x <= curr.Val {
            break
        }
        if prev.Val > curr.Val {
            if x >= prev.Val || x <= curr.Val {
                break
            }
        }
        prev, curr = curr, curr.Next
    }

    prev.Next = &Node{x, curr}
    return aNode
}

JavaScript

/**
 * // Definition for a Node.
 * function Node(val, next) {
 *     this.val = val;
 *     this.next = next;
 * };
 */

/**
 * @param {Node} head
 * @param {number} insertVal
 * @return {Node}
 */
var insert = function (head, insertVal) {
    if (!head) {
        const newNode = new Node(insertVal, null);
        newNode.next = newNode;
        return newNode;
    }

    let prev = head,
        curr = head.next;
    while (curr != head) {
        if (prev.val <= insertVal && insertVal <= curr.val) {
            break;
        }
        if (prev.val > curr.val && (insertVal >= prev.val || insertVal <= curr.val)) {
            break;
        }
        [prev, curr] = [curr, curr.next];
    }

    prev.next = new Node(insertVal, curr);
    return head;
};

Kotlin

/**
 * Definition for a Node.
 * class Node(var `val`: Int) {
 *     var next: Node? = null
 * }
 */

class Solution {
    fun insert(head: Node?, insertVal: Int): Node? {
        if (head == null) {
            val newNode: Node? = Node(insertVal)
            newNode?.next = newNode
            return newNode
        }

        var prev: Node? = head
        var curr: Node? = head?.next
        while (curr != head) {
            if (prev!!.`val` <= insertVal && insertVal <= curr!!.`val`) {
                break
            }
            if (prev!!.`val` > curr!!.`val`) {
                if (insertVal >= prev!!.`val` || insertVal <= curr!!.`val`) {
                    break
                }
            }

            prev = curr
            curr = curr?.next
        }

        val newNode: Node? = Node(insertVal)
        newNode?.next = curr
        prev?.next = newNode
        return head
    }
}

PHP

/**
 * Definition for a Node.
 * class Node {
 *     public $val = null;
 *     public $next = null;
 *     function __construct($val = 0) {
 *         $this->val = $val;
 *         $this->next = null;
 *     }
 * }
 */

class Solution
{
    /**
     * @param Node $root
     * @param Integer $insertVal
     * @return Node
     */
    function insert($head, $insertVal)
    {
        if (is_null($head)) {
            $node = new Node($insertVal);
            $node->next = $node;
            return $node;
        }

        $prev = $head;
        $curr = $head->next;
        while ($curr != $head) {
            if ($prev->val <= $insertVal && $insertVal <= $curr->val) {
                break;
            }
            if ($prev->val > $curr->val && ($insertVal >= $prev->val || $insertVal <= $curr->val)) {
                break;
            }
            $prev = $curr;
            $curr = $curr->next;
        }

        $node = new Node($insertVal);
        $node->next = $curr;
        $prev->next = $node;
        return $head;
    }
}

Python

"""
# Definition for a Node.
class Node:
    def __init__(self, val=None, next=None):
        self.val = val
        self.next = next
"""


class Solution:
    def insert(self, head: "Optional[Node]", insertVal: int) -> "Node":

        if head is None:
            newNode = Node(insertVal, None)
            newNode.next = newNode
            return newNode

        prev, curr = head, head.next
        while curr != head:
            if prev.val <= insertVal <= curr.val:
                # Case #1.
                break
            if prev.val > curr.val:
                # Case #2. where we locate the tail element
                # 'prev' points to the tail, i.e. the largest element!
                if insertVal >= prev.val or insertVal <= curr.val:
                    break
            prev, curr = curr, curr.next
        # Case #3.
        # did not insert the node in the loop
        prev.next = Node(insertVal, curr)
        return head

Rust


Swift

/**
 * Definition for a Node.
 * public class Node {
 *     public var val: Int
 *     public var next: Node?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.next = nil
 *     }
 * }
 */

class Solution {
    func insert(_ head: Node?, _ insertVal: Int) -> Node? {
        if head == nil {
            let newNode: Node? = Node(insertVal)
            newNode.next = newNode
            return newNode
        }

        var prev: Node? = head
        var curr: Node? = head?.next
        while curr != head {
            if prev!.val <= insertVal && insertVal <= curr!.val {
                break
            }
            if prev!.val > curr!.val {
                if insertVal >= prev!.val || insertVal <= curr!.val {
                    break
                }
            }
            prev = curr
            curr = curr?.next
        }

        var newNode: Node? = Node(insertVal)
        newNode.next = curr
        prev?.next = newNode
        return head
    }
}

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