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Cracking the Coding Interview - Fooo

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Solutions to Chapter 11 | System Design and Memory Limits<br />

9 while (in.hasNextInt()) {<br />

10 int n = in.nextInt();<br />

11 blocks[n / (bitfield.length * 8)]++;<br />

12 }<br />

13<br />

14 for (int i = 0; i < blocks.length; i++) {<br />

15 if (blocks[i] < bitfield.length * 8){<br />

16 /* if value < 2^20, <strong>the</strong>n at least 1 number is missing in<br />

17 * that section. */<br />

18 starting = i * bitfield.length * 8;<br />

19 break;<br />

20 }<br />

21 }<br />

22<br />

23 in = new Scanner(new FileReader(“input_file_q11_4.txt”));<br />

24 while (in.hasNextInt()) {<br />

25 int n = in.nextInt();<br />

26 /* If <strong>the</strong> number is inside <strong>the</strong> block that’s missing numbers,<br />

27 * we record it */<br />

28 if( n >= starting && n < starting + bitfield.length * 8){<br />

29 bitfield [(n-starting) / 8] |= 1

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