[LeetCode] 379. Design Phone Directory 设计电话目录
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);
设计一个电话目录有3个功能，get: 给出一个没有使用的电话号码，check: 查验一个电话号码是否被使用了，release: 释放一个电话号码，使其能被从新使用。
1、哈希表：只需要存贮一个哈希表，所以空间复杂度理论上是O(n)（实际上存储空间应该比这个大，因为涉及到散列和解决冲突）。
PhoneDirectory(int maxNumbers)：时间复杂度是O(n)。
int get()：时间复杂度是O(1)。
bool check(int number)：时间复杂度是O(1)。
void release(int number)：时间复杂度是O(1)。
2、二叉搜索树：只需要存贮一棵二叉搜索树，所以空间复杂度是O(n)。
PhoneDirectory(int maxNumbers)：时间复杂度是O(nlogn)。
int get()：时间复杂度是O(logn)。
bool check(int number)：时间复杂度是O(logn)。
void release(int number)：时间复杂度是O(logn)。
3、数组：存储一个表示电话号码的数组，以及一个表示某个电话号码是否可用的数组，以及一个当前可用的电话号码个数。所以空间复杂度是O(2n)。
PhoneDirectory(int maxNumbers)：时间复杂度是O(n)。
int get()：时间复杂度是O(1)。
bool check(int number)：时间复杂度是O(1)。
void release(int number)：时间复杂度是O(1)。
通过对比可以发现，1和3的时间复杂度都较低，但是由于3中每个函数中同时要对两个数组进行操作，所以运行时间还是要略慢于1。这是不是也说明C++系统中实现的散列函数还是不错的，解决冲突并没有花费太多的时间？
解法1: hash table, 不好
解法2: binary tree, 不好
解法3: 数组
解法4: n-ary tree
G家：Phone number assignment
Java:
public class PhoneDirectory {
int max;
HashSet<Integer> set;
LinkedList<Integer> queue;
/** Initialize your data structure here
@param maxNumbers - The maximum numbers that can be stored in the phone directory. */
public PhoneDirectory(int maxNumbers) {
set = new HashSet<Integer>();
queue = new LinkedList<Integer>();
for(int i=0; i<maxNumbers; i++){
queue.offer(i);
}
max=maxNumbers-1;
}
/** Provide a number which is not assigned to anyone.
@return - Return an available number. Return -1 if none is available. */
public int get() {
if(queue.isEmpty())
return -1;
int e = queue.poll();
set.add(e);
return e;
}
/** Check if a number is available or not. */
public boolean check(int number) {
return !set.contains(number) && number<=max;
}
/** Recycle or release a number. */
public void release(int number) {
if(set.contains(number)){
set.remove(number);
queue.offer(number);
}
}
}
Python:
# init: Time: O(n), Space: O(n)
# get: Time: O(1), Space: O(1)
# check: Time: O(1), Space: O(1)
# release: Time: O(1), Space: O(1)
class PhoneDirectory(object):
def __init__(self, maxNumbers):
"""
Initialize your data structure here
@param maxNumbers - The maximum numbers that can be stored in the phone directory.
:type maxNumbers: int
"""
self.__curr = 0
self.__numbers = range(maxNumbers)
self.__used = [False] * maxNumbers
def get(self):
"""
Provide a number which is not assigned to anyone.
@return - Return an available number. Return -1 if none is available.
:rtype: int
"""
if self.__curr == len(self.__numbers):
return -1
number = self.__numbers[self.__curr]
self.__curr += 1
self.__used[number] = True
return number
def check(self, number):
"""
Check if a number is available or not.
:type number: int
:rtype: bool
"""
return 0 <= number < len(self.__numbers) and \
not self.__used[number]
def release(self, number):
"""
Recycle or release a number.
:type number: int
:rtype: void
"""
if not 0 <= number < len(self.__numbers) or \
not self.__used[number]:
return
self.__used[number] = False
self.__curr -= 1
self.__numbers[self.__curr] = number
# Your PhoneDirectory object will be instantiated and called as such:
# obj = PhoneDirectory(maxNumbers)
# param_1 = obj.get()
# param_2 = obj.check(number)
# obj.release(number)
C++: Hash table
class PhoneDirectory {
public:
/** Initialize your data structure here
@param maxNumbers - The maximum numbers that can be stored in the phone directory. */
PhoneDirectory(int maxNumbers) {
for (int i = 0; i < maxNumbers; ++i) {
hash.insert(i);
}
}
/** Provide a number which is not assigned to anyone.
@return - Return an available number. Return -1 if none is available. */
int get() {
if (hash.empty()) {
return -1;
}
int ret = *(hash.begin());
hash.erase(hash.begin());
return ret;
}
/** Check if a number is available or not. */
bool check(int number) {
return hash.count(number) > 0;
}
/** Recycle or release a number. */
void release(int number) {
hash.insert(number);
}
private:
unordered_set<int> hash;
};
/**
* 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);
*/
C++: binary tree
class PhoneDirectory {
public:
/** Initialize your data structure here
@param maxNumbers - The maximum numbers that can be stored in the phone directory. */
PhoneDirectory(int maxNumbers) {
for (int i = 0; i < maxNumbers; ++i) {
st.insert(i);
}
}
/** Provide a number which is not assigned to anyone.
@return - Return an available number. Return -1 if none is available. */
int get() {
if (st.empty()) {
return -1;
}
int ret = *st.begin();
st.erase(st.begin());
return ret;
}
/** Check if a number is available or not. */
bool check(int number) {
return st.find(number) != st.end();
}
/** Recycle or release a number. */
void release(int number) {
st.insert(number);
}
private:
set<int> st;
};
/**
* 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);
*/
C++: array
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;
numbers.resize(n);
for(int i = 0; i < n; ++i) {
numbers[i] = i;
}
is_available.resize(n, true);
}
/** 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 num = numbers[--n];
is_available[num] = false;
return num;
}
/** Check if a number is available or not. */
bool check(int number) {
return is_available[number];
}
/** Recycle or release a number. */
void release(int number) {
if(is_available[number]) { // already released
return;
}
numbers[n++] = number;
is_available[number] = true;
}
private:
vector<int> numbers; // the number array
vector<bool> is_available; // indicate whether the number is available
int n; // the current available numbers
};
/**
* 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);
*/
C++:
class PhoneDirectory {
public:
/** Initialize your data structure here
@param maxNumbers - The maximum numbers that can be stored in the phone directory. */
PhoneDirectory(int maxNumbers) {
max_num = maxNumbers;
next = idx = 0;
recycle.resize(max_num);
flag.resize(max_num, 1);
}
/** Provide a number which is not assigned to anyone.
@return - Return an available number. Return -1 if none is available. */
int get() {
if (next == max_num && idx <= 0) return -1;
if (idx > 0) {
int t = recycle[--idx];
flag[t] = 0;
return t;
}
flag[next] = false;
return next++;
}
/** Check if a number is available or not. */
bool check(int number) {
return number >= 0 && number < max_num && flag[number];
}
/** Recycle or release a number. */
void release(int number) {
if (number >= 0 && number < max_num && !flag[number]) {
recycle[idx++] = number;
flag[number] = 1;
}
}
private:
int max_num, next, idx;
vector<int> recycle, flag;
};
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