《Cracking the Coding Interview》——第10章：可扩展性和存储空间限制——题目3

2014-04-24 21:06

题目：假设你有40亿个不重复的非负整数存在一个文件里，1GB的内存。你要如何找出一个文件中不包含的整数呢？如果是10亿个整数，10MB内存又该如何？

解法：前者内存充足，可以直接用一个位向量来标记整数的出现情况，然后顺序扫描找出第一个没出现过的数字。后者内存不足，可以把整数范围分成多段，每次扫描整个文件，判断该段中的所有整数是否都出现过了。如果有没出现的，答案就找到了，否则继续扫描下一段。

代码：

// 10.3 Assume you have a list of 4 billion non-negative integers, you're given 1GB memory. Try to find out a number that is not contained in the number list.
// Answer:
//    1GB means 8Gbit, thus we can use 4 billion bit vector, of actual size 500MB. Scan the list for one pass and mark the existent as '1'.
//    Scan the bit vector to find the first '0' number.
// Extension: if you have only 10MB memory, what's it gonna be then?
// Answer:
//    10MB means 80Mbit, thus we have to scan the list for multiple time. Every time the bit vector covers only part of the range.
//    Such as 0~999, 1000~199, ..., 3999999000~3999999999
//    For each interval, we scan the all numbers and check if there is missing element in the bit vector. If found, return it. Otherwise go on until the last interval.
//    the length of the interval should be carefully chosen, to minimize total number of scannings.
int main()
{
    return 0;
}