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);分析
1 一个set放released, 一个int记录当前已达数字（++--）
2 一个set装满所有available，加加减减.cons 初始要O(N)
3 java BitSet 和 index。 index就是当前数，release 就clear and set index. get 就 return index and set index
4 segment tree
方法1:
release要有判断，max不可达
class PhoneDirectory:
def __init__(self, maxNumbers: int):
"""
Initialize your data structure here
@param maxNumbers - The maximum numbers that can be stored in the phone directory.
"""
self.next = 0
self.released = set()
self.max = maxNumbers
def get(self) -> int:
"""
Provide a number which is not assigned to anyone.
@return - Return an available number. Return -1 if none is available.
"""
res = 0
if self.next < self.max:
res = self.next
self.next += 1
return res
if self.released:
return self.released.pop()
return -1
def check(self, number: int) -> bool:
"""
Check if a number is available or not.
"""
if number in self.released or self.next <= number < self.max :
return True
return False
def release(self, number: int) -> None:
"""
Recycle or release a number.
"""
if number < self.next: #记得要判断这里，而且不是<max
self.released.add(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)方法2：
class PhoneDirectory:
def __init__(self, maxNumbers: int):
"""
Initialize your data structure here
@param maxNumbers - The maximum numbers that can be stored in the phone directory.
"""
self.avail = set(range(maxNumbers))
def get(self) -> int:
"""
Provide a number which is not assigned to anyone.
@return - Return an available number. Return -1 if none is available.
"""
if self.avail:
return self.avail.pop()
return -1
def check(self, number: int) -> bool:
"""
Check if a number is available or not.
"""
return number in self.avail
def release(self, number: int) -> None:
"""
Recycle or release a number.
"""
self.avail.add(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)3 bitset
Space is much efficient: O(c), wouldn’t say it is O(1) because we still need max_size number of bits
这里bit的长度 = max number,某个数字就是设置这个位上的bit
nextClearBit(num)注意此处需要参数
class PhoneDirectory {
/** Initialize your data structure here
@param maxNumbers - The maximum numbers that can be stored in the phone directory. */
BitSet bs;
final int max;
int nextAvail;
public PhoneDirectory(int maxNumbers) {
bs = new BitSet(maxNumbers);
max = maxNumbers;
nextAvail = 0;
}
/** Provide a number which is not assigned to anyone.
@return - Return an available number. Return -1 if none is available. */
public int get() {
if (nextAvail >= max) {
return -1;
}
int res = nextAvail;
bs.set(res);
nextAvail = bs.nextClearBit(res);//注意nextClearBit用法，需要传入num
return res;
}
/** Check if a number is available or not. */
public boolean check(int number) {
return number < max && !bs.get(number);
}
/** Recycle or release a number. */
public void release(int number) {
if(number >= 0 && number < max){
bs.clear(number);
}
nextAvail = Math.min(number,nextAvail);
}
}
/**
* Your PhoneDirectory object will be instantiated and called as such:
* PhoneDirectory obj = new PhoneDirectory(maxNumbers);
* int param_1 = obj.get();
* boolean param_2 = obj.check(number);
* obj.release(number);
*/4 segment Tree
O(log n) time in allocate and release
总共2*max个节点，max个叶节点》max，max个内节点 《max。线段树是Bool arr
get取第一个超过max的未赋值，也就是叶节点。 每次使用叶节点或者释放，都要更新内节点。pushup
class PhoneDirectory:
def __init__(self, maxNumbers: int):
"""
Initialize your data structure here
@param maxNumbers - The maximum numbers that can be stored in the phone directory.
"""
self.max = maxNumbers
self.segTree = [True]*(self.max << 1)
def pushup(self, i) -> None:
i >>=1
while i > 0:
self.segTree[i] = self.segTree[i<<1] or self.segTree[i<<1|1]
i>>=1
def get(self) -> int:
"""
Provide a number which is not assigned to anyone.
@return - Return an available number. Return -1 if none is available.
"""
if not self.segTree[1]:
return -1
i = 1
while i < self.max:
if (i << 1) < (self.max << 1) and self.segTree[i<<1]:
i = i<<1
if i<< 1|1 < self.max << 1 and self.segTree[i<<1|1]:
i = i << 1|1
self.segTree[i] = False
res = i - self.max
self.pushup(i)
return res
def check(self, number: int) -> bool:
"""
Check if a number is available or not.
"""
if 0<=number < self.max:
return self.segTree[number + self.max]
return False
def release(self, number: int) -> None:
"""
Recycle or release a number.
"""
i = number+self.max
self.segTree[i] = True
self.pushup(i)
# 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)Last updated
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