我的面试问题是这样的:
给定一个包含40亿个整数的输入文件，提供一种算法来生成一个文件中不包含的整数。假设您有1gb内存。如果你只有10mb的内存，你会怎么做。
我的分析:
文件大小为4×109×4 bytes = 16gb。
我们可以进行外部排序，从而知道整数的范围。
我的问题是，在已排序的大整数集中检测缺失整数的最佳方法是什么?
我的理解(看完所有答案后):
假设我们讨论的是32位整数，有232 = 4*109个不同的整数。
情况1:我们有1gb = 1 * 109 * 8位= 80亿位内存。
解决方案:
如果我们用一位表示一个不同的整数，这就足够了。我们不需要排序。
实现:
int radix = 8;
byte[] bitfield = new byte[0xffffffff/radix];
void F() throws FileNotFoundException{
Scanner in = new Scanner(new FileReader("a.txt"));
while(in.hasNextInt()){
int n = in.nextInt();
bitfield[n/radix] |= (1 << (n%radix));
}
for(int i = 0; i< bitfield.lenght; i++){
for(int j =0; j<radix; j++){
if( (bitfield[i] & (1<<j)) == 0) System.out.print(i*radix+j);
}
}
}
情形二:10mb内存= 10 * 106 * 8bits = 8000万bits
Solution:
For all possible 16-bit prefixes, there are 216 number of
integers = 65536, we need 216 * 4 * 8 = 2 million bits. We need build 65536 buckets. For each bucket, we need 4 bytes holding all possibilities because the worst case is all the 4 billion integers belong to the same bucket.
Build the counter of each bucket through the first pass through the file.
Scan the buckets, find the first one who has less than 65536 hit.
Build new buckets whose high 16-bit prefixes are we found in step2
through second pass of the file
Scan the buckets built in step3, find the first bucket which doesnt
have a hit.
The code is very similar to above one.
结论:
我们通过增加文件传递来减少内存。
为那些迟到的人澄清一下:这个问题，正如所问的，并不是说有一个整数没有包含在文件中——至少大多数人不是这么理解的。不过，评论线程中的许多评论都是关于任务的变化。不幸的是，把它引入评论线程的评论后来被它的作者删除了，所以现在看起来它的孤儿回复只是误解了一切。这很让人困惑，抱歉。
The simplest approach is to find the minimum number in the file, and return 1 less than that. This uses O(1) storage, and O(n) time for a file of n numbers. However, it will fail if number range is limited, which could make min-1 not-a-number.
The simple and straightforward method of using a bitmap has already been mentioned. That method uses O(n) time and storage.
A 2-pass method with 2^16 counting-buckets has also been mentioned. It reads 2*n integers, so uses O(n) time and O(1) storage, but it cannot handle datasets with more than 2^16 numbers. However, it's easily extended to (eg) 2^60 64-bit integers by running 4 passes instead of 2, and easily adapted to using tiny memory by using only as many bins as fit in memory and increasing the number of passes correspondingly, in which case run time is no longer O(n) but instead is O(n*log n).
The method of XOR'ing all the numbers together, mentioned so far by rfrankel and at length by ircmaxell answers the question asked in stackoverflow#35185, as ltn100 pointed out. It uses O(1) storage and O(n) run time. If for the moment we assume 32-bit integers, XOR has a 7% probability of producing a distinct number. Rationale: given ~ 4G distinct numbers XOR'd together, and ca. 300M not in file, the number of set bits in each bit position has equal chance of being odd or even. Thus, 2^32 numbers have equal likelihood of arising as the XOR result, of which 93% are already in file. Note that if the numbers in file aren't all distinct, the XOR method's probability of success rises.
我认为这是一个已解决的问题(见上文)，但还有一个有趣的情况需要记住，因为它可能会被问到:
如果恰好有4,294,967,295(2^32 - 1)个没有重复的32位整数，因此只有一个缺失，有一个简单的解决方案。
从0开始计算运行总数，对于文件中的每个整数，将该整数加上32位溢出(实际上，runningTotal = (runningTotal + nextInteger) % 4294967296)。一旦完成，将4294967296/2加到运行总数中，同样是32位溢出。用4294967296减去这个，结果就是缺少的整数。
“只缺少一个整数”的问题只需运行一次就可以解决，并且只有64位RAM专用于数据(运行总数为32位，读入下一个整数为32位)。
推论:如果我们不关心整数结果必须有多少位，那么更通用的规范非常容易匹配。我们只是生成一个足够大的整数，它不能包含在我们给定的文件中。同样，这只占用极小的RAM。请参阅伪代码。
# Grab the file size
fseek(fp, 0L, SEEK_END);
sz = ftell(fp);
# Print a '2' for every bit of the file.
for (c=0; c<sz; c++) {
for (b=0; b<4; b++) {
print "2";
}
}
If they are 32-bit integers (likely from the choice of ~4 billion numbers close to 232), your list of 4 billion numbers will take up at most 93% of the possible integers (4 * 109 / (232) ). So if you create a bit-array of 232 bits with each bit initialized to zero (which will take up 229 bytes ~ 500 MB of RAM; remember a byte = 23 bits = 8 bits), read through your integer list and for each int set the corresponding bit-array element from 0 to 1; and then read through your bit-array and return the first bit that's still 0.
In the case where you have less RAM (~10 MB), this solution needs to be slightly modified. 10 MB ~ 83886080 bits is still enough to do a bit-array for all numbers between 0 and 83886079. So you could read through your list of ints; and only record #s that are between 0 and 83886079 in your bit array. If the numbers are randomly distributed; with overwhelming probability (it differs by 100% by about 10-2592069) you will find a missing int). In fact, if you only choose numbers 1 to 2048 (with only 256 bytes of RAM) you'd still find a missing number an overwhelming percentage (99.99999999999999999999999999999999999999999999999999999999999995%) of the time.
但我们假设不是有40亿个数字;你有232 - 1这样的数字和不到10mb的RAM;所以任何小范围的整数都只有很小的可能性不包含这个数字。
如果保证列表中的每个int都是唯一的，那么可以将这些数字相加，并减去一个#，再减去完整的和(½)(232)(232 - 1)= 9223372034707292160，以找到缺少的int。但是，如果出现了两次int，则此方法将失败。
However, you can always divide and conquer. A naive method, would be to read through the array and count the number of numbers that are in the first half (0 to 231-1) and second half (231, 232). Then pick the range with fewer numbers and repeat dividing that range in half. (Say if there were two less number in (231, 232) then your next search would count the numbers in the range (231, 3*230-1), (3*230, 232). Keep repeating until you find a range with zero numbers and you have your answer. Should take O(lg N) ~ 32 reads through the array.
这种方法效率很低。我们在每一步中只使用两个整数(或者大约8字节的RAM和一个4字节(32位)整数)。更好的方法是将其划分为sqrt(232) = 216 = 65536个箱子，每个箱子中有65536个数字。每个bin需要4个字节来存储它的计数，因此需要218字节= 256 kB。因此，bin 0为(0 ~ 65535=216-1)，bin 1为(216=65536 ~ 2*216-1=131071)，bin 2为(2*216=131072 ~ 3*216-1=196607)。在python中，你会有这样的代码:
import numpy as np
nums_in_bin = np.zeros(65536, dtype=np.uint32)
for N in four_billion_int_array:
nums_in_bin[N // 65536] += 1
for bin_num, bin_count in enumerate(nums_in_bin):
if bin_count < 65536:
break # we have found an incomplete bin with missing ints (bin_num)
通读~ 40亿整数列表;然后计算216个容器中每个容器中有多少int，并找到一个不包含65536个数字的incomplete_bin。然后你再读一遍40亿的整数列表;但这次只注意整数在这个范围内;当你找到他们的时候，你会有点抓狂。
del nums_in_bin # allow gc to free old 256kB array
from bitarray import bitarray
my_bit_array = bitarray(65536) # 32 kB
my_bit_array.setall(0)
for N in four_billion_int_array:
if N // 65536 == bin_num:
my_bit_array[N % 65536] = 1
for i, bit in enumerate(my_bit_array):
if not bit:
print bin_num*65536 + i
break
好的，这并没有经过充分的思考，因为它假设文件中的整数遵循某种静态分布。显然他们不需要这样做，但即使这样，也应该试试这个:
有≈43亿个32位整数。我们不知道它们在文件中是如何分布的，但最糟糕的情况是具有最高香农熵的情况:均匀分布。在这种情况下，任何一个整数不出现在文件中的概率为
((2³²-1)/2³²)⁰⁰⁰⁰≈.4
The lower the Shannon entropy, the higher this probability gets on the average, but even for this worst case we have a chance of 90% to find a nonoccurring number after 5 guesses with random integers. Just create such numbers with a pseudorandom generator, store them in a list. Then read int after int and compare it to all of your guesses. When there's a match, remove this list entry. After having been through all of the file, chances are you will have more than one guess left. Use any of them. In the rare (10% even at worst case) event of no guess remaining, get a new set of random integers, perhaps more this time (10->99%).
内存消耗:几十个字节，复杂度:O(n)，开销:neclectable，因为大部分时间将花费在不可避免的硬盘访问上，而不是比较int类型。
当我们不假设静态分布时，实际最坏的情况是每个整数都出现最大值。曾经，因为那时只有
1 - 4000000000/2³²≈6%
所有的整数都不会出现在文件中。因此，您需要更多的猜测，但这仍然不会消耗大量的内存。
