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。
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";
}
}
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.
