Design Phone Directory

NicolasHe 发布于2019-08-14 17:26 / 2425人阅读

摘要：题目链接直接用一个，结果了看了加了个，不过感觉没什么必要加，反正保存的都一样，只是的时间大于，用可以保证。看了题目条件是可以随便返回一个值，但是不让这么做。很无语啊如果这道题要求要求的是的，那就和一样了。

Design Phone Directory

题目链接：https://leetcode.com/problems...

直接用一个set，结果tle了= =

</>复制代码 
public class PhoneDirectory {
    /** Initialize your data structure here
        @param maxNumbers - The maximum numbers that can be stored in the phone directory. */
    Set set = new HashSet();
    int maxNum;
    public PhoneDirectory(int maxNumbers) {
        for(int i = 0; i < maxNumbers; i++) set.add(i);
        maxNum = maxNumbers;
    }
    
    /** Provide a number which is not assigned to anyone.
        @return - Return an available number. Return -1 if none is available. */
    public int get() {
        // no available numbers left
        if(set.size() == 0) return -1;
        // return 1 number
        else {
            int number = set.iterator().next();
            set.remove(number);
            return number;
        }
    }
    
    /** Check if a number is available or not. */
    public boolean check(int number) {
        return set.contains(number);
    }
    
    /** Recycle or release a number. */
    public void release(int number) {
        if(number >= 0 && number < maxNum) set.add(number);
    }
}

看了discussion加了个queue，不过感觉没什么必要加q，反正保存的都一样，只是set.iterator().next()的时间大于O(1)，用q可以保证O(1)。看了题目条件是get可以随便返回一个值，但是test case不让这么做。很无语啊

</>复制代码 
public class PhoneDirectory {
    /** Initialize your data structure here
        @param maxNumbers - The maximum numbers that can be stored in the phone directory. */
    Set set = new HashSet();
    int maxNum;
    Queue available = new LinkedList();
    public PhoneDirectory(int maxNumbers) {
        for(int i = 0; i < maxNumbers; i++) available.add(i);
        maxNum = maxNumbers;
    }
    
    /** Provide a number which is not assigned to anyone.
        @return - Return an available number. Return -1 if none is available. */
    public int get() {
        // no available numbers left
        if(available.size() == 0) return -1;
        // return 1 number
        else {
            int number = available.poll();
            set.add(number);
            return number;
        }
    }
    
    /** Check if a number is available or not. */
    public boolean check(int number) {
        return !set.contains(number);
    }
    
    /** Recycle or release a number. */
    public void release(int number) {
        if(number >= 0 && number < maxNum) {
            if(set.remove(number)) available.add(number);
        }
    }
}

如果这道题要求get要求的是random的，那就和380. Insert Delete GetRandom O(1)一样了。
https://segmentfault.com/a/11...

GPU云服务器云服务器 Directory Phone sae中使用directory windows phone 8

文章版权归作者所有，未经允许请勿转载,若此文章存在违规行为，您可以联系管理员删除。

转载请注明本文地址：https://www.ucloud.cn/yun/66633.html

[LeetCode] Design Phone Directory

Problem Design a Phone Directory which supports the following operations: get: Provide a number which is not assigned to anyone.check: Check if a number is available or not.release: Recycle or release...

wangbinke 2019-08-14 15:57 评论0 收藏0
[LeetCode] 588. Design In-Memory File System

Problem Design an in-memory file system to simulate the following functions: ls: Given a path in string format. If it is a file path, return a list that only contains this files name. If it is a direc...

SHERlocked93 2019-08-16 13:57 评论0 收藏0
React框架Umi实战(2)整合dva开发后台管理系统

摘要：官方推荐结合使用更配哦，其实他们都是同一位开发者开发的，属于阿里内部开源框架。但是名字必须为，不然不能自动注册。只有一个的话，可以不用建目录。可能是我理解有误。 umi官方推荐结合dva使用更配哦，其实他们都是同一位开发者开发的，属于阿里内部开源框架。 1 修改.umirc.js，开启dva支持 // ref: https://umijs.org/config/ export def...

beita 2019-08-23 14:39 评论0 收藏0