[LeetCode] 2336. Smallest Number in Infinite Set

Posted 2023-04-26 CNoodle

You have a set which contains all positive integers [1, 2, 3, 4, 5, ...].

Implement the SmallestInfiniteSet class:

• SmallestInfiniteSet() Initializes the SmallestInfiniteSet object to contain all positive integers.
• int popSmallest() Removes and returns the smallest integer contained in the infinite set.
• void addBack(int num) Adds a positive integer num back into the infinite set, if it is not already in the infinite set.

Example 1:

Input
["SmallestInfiniteSet", "addBack", "popSmallest", "popSmallest", "popSmallest", "addBack", "popSmallest", "popSmallest", "popSmallest"]
[[], [2], [], [], [], [1], [], [], []]
Output
[null, null, 1, 2, 3, null, 1, 4, 5]

Explanation
SmallestInfiniteSet smallestInfiniteSet = new SmallestInfiniteSet();
smallestInfiniteSet.addBack(2);    // 2 is already in the set, so no change is made.
smallestInfiniteSet.popSmallest(); // return 1, since 1 is the smallest number, and remove it from the set.
smallestInfiniteSet.popSmallest(); // return 2, and remove it from the set.
smallestInfiniteSet.popSmallest(); // return 3, and remove it from the set.
smallestInfiniteSet.addBack(1);    // 1 is added back to the set.
smallestInfiniteSet.popSmallest(); // return 1, since 1 was added back to the set and
                                   // is the smallest number, and remove it from the set.
smallestInfiniteSet.popSmallest(); // return 4, and remove it from the set.
smallestInfiniteSet.popSmallest(); // return 5, and remove it from the set.

Constraints:

• 1 <= num <= 1000
• At most 1000 calls will be made in total to popSmallest and addBack.

无限集中的最小数字。

现有一个包含所有正整数的集合 [1, 2, 3, 4, 5, ...] 。

实现 SmallestInfiniteSet 类：

SmallestInfiniteSet() 初始化 SmallestInfiniteSet 对象以包含 所有 正整数。
int popSmallest() 移除 并返回该无限集中的最小整数。
void addBack(int num) 如果正整数 num 不 存在于无限集中，则将一个 num 添加 到该无限集中。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/smallest-number-in-infinite-set
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路是最小堆 + hashset。初始化的时候，我们将 1 - 1000 这些数字都分别放入最小堆和 hashset 中。

pop() 很好判断，就是从最小堆中弹出一个元素即可，同时在 hashset 中也移除这个元素。

addBack() 需要判断最小堆中是否已经存在这个数字了，如果不存在才加回去。注意 testcase 中对于 add 操作，给的数字不一定是已经弹出的元素。

时间O(nlogn)

空间O(n)

Java实现

 class SmallestInfiniteSet 
     PriorityQueue<Integer> queue;
     Set<Integer> set;
 
     public SmallestInfiniteSet() 
         queue = new PriorityQueue<>();
         set = new HashSet<>();
         for (int i = 1; i <= 1000; i++) 
             queue.offer(i);
             set.add(i);
         
     
     
     public int popSmallest() 
         if (!queue.isEmpty()) 
             set.remove(queue.peek());
             return queue.poll();
         
         return -1;
     
     
     public void addBack(int num) 
         if (!set.contains(num)) 
             set.add(num);
             queue.offer(num);
         
     
 
 
 /**
  * Your SmallestInfiniteSet object will be instantiated and called as such:
  * SmallestInfiniteSet obj = new SmallestInfiniteSet();
  * int param_1 = obj.popSmallest();
  * obj.addBack(num);
  */

 

LeetCode 题目总结

Leetcode 230. Kth Smallest Element in a BST

题目链接

https://leetcode.com/problems/kth-smallest-element-in-a-bst/description/

题目描述

Given a binary search tree, write a function kthSmallest to find the kth smallest element in it.

Note:
You may assume k is always valid, 1 ≤ k ≤ BST‘s total elements.

Example 1:

Input: root = [3,1,4,null,2], k = 1
   3
  /  1   4
     2
Output: 1

Example 2:

Input: root = [5,3,6,2,4,null,null,1], k = 3
       5
      /      3   6
    /    2   4
  /
 1
Output: 3

Follow up:
What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?

题解

最直接的方法就是中序遍历，得到有序数组，然后取得第k个值；
也可以使用二分法，因为平衡二叉树以根节点分为左右两部分，我们可以统计左边节点的个数，然后与k值相比较，如果大于k值，说明第k个值在左子树上，否则在当前节点或者右子树。递归遍历即可。

代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int kthSmallest(TreeNode root, int k) {
        int count = countNodes(root.left);
        if (k <= count) {
            return kthSmallest(root.left, k);
        } else if (k > count + 1) {
            return kthSmallest(root.right, k - 1 - count);
        }
        return root.val;
    }
    
    public int countNodes(TreeNode n) {
        if (n == null) { return 0; }
        return 1 + countNodes(n.left) + countNodes(n.right);
    }
}