[leetcode] 379. Design Phone Directory 解题报告

题目链接: https://leetcode.com/problems/design-phone-directory/

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 a number.

Example:

// Init a phone directory containing a total of 3 numbers: 0, 1, and 2.
PhoneDirectory directory = new PhoneDirectory(3);// It can return any available phone number. Here we assume it returns 0.
directory.get();// Assume it returns 1.
directory.get();// The number 2 is available, so return true.
directory.check(2);// It returns 2, the only number that is left.
directory.get();// The number 2 is no longer available, so return false.
directory.check(2);// Release number 2 back to the pool.
directory.release(2);// Number 2 is available again, return true.
directory.check(2);

思路: 这题的对C++时间卡的太严了, 基本上只要涉及到任何高级数据结构就会超时, 只能利用c风格的代码, 并且所有操作除初始化以外都是O(1)才可以过.

可以利用两个数组和一个当前剩余号码计数, 一个数组前n个保存当前可用的号码, 另一个数组映射每一个号码是否可用, 还有一个计数是当前第一个数组前n个是可用的.

代码如下:

class PhoneDirectory {
public:/** Initialize your data structure here@param maxNumbers - The maximum numbers that can be stored in the phone directory. */PhoneDirectory(int maxNumbers){n = maxNumbers;available = new int[n];isAvailable = new int[n]; for(int i = 0; i < n; i++)available[i] = i, isAvailable[i] = 1;}~PhoneDirectory(){delete [] available;delete [] isAvailable;}/** Provide a number which is not assigned to anyone.@return - Return an available number. Return -1 if none is available. */int get() {if(n <= 0) return -1;int ans = available[--n];isAvailable[ans] = 0;return ans;}/** Check if a number is available or not. */bool check(int number) {return isAvailable[number];}/** Recycle or release a number. */void release(int number) {if(isAvailable[number]) return;available[n++] = number;isAvailable[number] = 1;}
private:int *available, *isAvailable;int n;
};/*** Your PhoneDirectory object will be instantiated and called as such:* PhoneDirectory obj = new PhoneDirectory(maxNumbers);* int param_1 = obj.get();* bool param_2 = obj.check(number);* obj.release(number);*/

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