9.10 Extensibility and Storage Limits (ii)--given an input file, contains 4 billion non-negative integers. Produces an integer that is not in the file. Memory Limit: 1GB

/**
* Function: Given an input file, contains 4 billion non-negative integers. Produces an integer that is not in the file. Memory Limit: 1GB
* Advanced: Memory limit 10MB.

*/

/** * Idea: *  * 1) Create bit vectors that contain 40个亿个 bits.    the bit vector (bv,bit vector) is actually an array that uses an integer (or another data type) array to store the Boolean values compactly. Each integer can store a string of 32-bit or Boolean values. * 2) Initialize all elements of BV to 0. * 3) Scan all the numbers in the file (num) and call Bv.set (num,1). * 4) Next, start scanning BV again from index 0. * 5) returns the first index with a value of 0. * * *  value is determined: * 1) 2^32-->40 billion different integers, an integer of 4 bytes. * 2) 2^30 byte-->1gb-->2^33bit-->80 million bits. Each bit maps a different integer that can store more than 8 billion different integers. */long numberofints= ((long) integer.max_value) +1;byte[] bitfield=new byte[(int) (NUMBEROFINTS/8)];p ublic void Findopennumber () throws Filenotfoundexception{scanner in=new Scanner (New FileReader ("F:\\file.txt")); while ( In.hasnext ()) {int n=in.nextint ();/*  use the OR operator to set the nth bit of a byte to find the corresponding number in the Bitfield. * (for example, 10 (decimal) will correspond to the 2nd bit of index 2 in the byte array) */bitfield[n/8]= (byte) (1<< (n%8));} for (int i=0;i<bitfield.length;i++) {for (int j=0;j<8;j++) {/* * *) retrieves each bit of each byte. When a bit is found to be 0 o'clock, the corresponding value is found. */if ((bitfield[i]& (1<<j)) ==0) {System.out.println (i*8+j); return;}}}

Advanced: 10MB

/** * Idea: To scan a data set two times, you can find an integer that is not in the file. You can divide all integers into chunks of the same size. * First Scan array: determines the number of elements per array. * Second scan bit vector: Determines the number less in the range. * * Bitsize:: The size of each block range for the first scan. * Blocknum: The number of blocks during the first scan. * * value is determined: * 1) 10mb-->2^23byte. An integer of 4 bytes, so that it contains a maximum of 2^21 elements of an array. * 2) bitsize= (2^32/blocknum) <=2^21, so, blocknum>=2^11. * 2^11<=bitsize<=2^26. * Under these conditions, the more the median value is chosen, the less memory is being sued at any time. */int bitsize=1048576;int blocknum=4096;byte[] bitfield2=new byte[bitsize/8];int[] blocks=new Int[blockNum];p ublic void FindOpenNumber2 () throws Filenotfoundexception{scanner in=new Scanner (New FileReader ("F:\\file.txt")); int Starting=-1;while (In.hasnext ()) {int n=in.nextint (); blocks[n/bitfield2.length*8]++;} If the for (int i=0;i<blocks.length;i++) {/* * is less than, it indicates that there is at least one number missing in the Block */if (blocks[i]<bitfield2.length*8) {starting=i* Bitfield2.length*8;break;}} In =new Scanner (new FileReader ("F:\\file.txt")), while (In.hasnext ()) {int n=in.nextint (); if (n>=starting&& n<starting+bitfield2.length*8) {bitfield2[(n-starting)/8]= (Byte) (1<< ((n-starting)%8));}} for (iNT i=0;i<bitfield2.length;i++) {for (int j=0;j<8;j++) {/* * * Retrieve each bit of each byte. When a bit is found to be 0 o'clock, the corresponding value is found. */if ((bitfield2[i]& (1<<j)) ==0) {System.out.println (i*8+j+starting); return;}}}

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9.10 Extensibility and Storage Limits (ii)--given an input file, contains 4 billion non-negative integers. Produces an integer that is not in the file. Memory Limit: 1GB