74 use bitset to output repeated elements in the array

[Link to this article]

Http://www.cnblogs.com/hellogiser/p/using-bitset-to-print-duplicate-elements-of-array.html

[Question]

An array has l elements in the range of 0 to n, where n is <32000, but the exact size of N is unknown. L there are several repeated elements in each element, with only 4 kb of memory. How can I output repeated elements?

[Analysis]

Since the value of the array is between a range [), we naturally want to use bitset for processing. If bitset is used, one integer can be expressed as one bit. 1 byte can represent 8 different integers, and 4 kb can represent 32 KB different integers. Because 32 KB = 32*1024> 32000, the 4 kb memory is sufficient to indicate 32000 different numbers. The core is the access to one integer in the bitset. Open an array that can store N/32 + 1 int numbers, then the number X is stored in the X % 32 bit of the array [x/32] integer.

Instant MessagingI = x/32, j = x % 32, array [I] | = (1 <j)

[Code]

C ++ code

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// 74_bitset.cpp: defines the entry point for the console application. // # Include "stdafx. H" # Include "iostream" # Include <ctime> # Include <algorithm> Using namespace STD; Class bitset { Public: Bitset (INT range) { // [0, range) M_nrange = range; M_nlength = m_nrange/32 + 1; Bitset = new int [m_nlength]; // Init all with 0 For (INT I = 0; I <m_nlength; I ++) { Bitset [I] = 0; } } ~ Bitset () { Delete [] bitset; } Void set (INT number) { Int I = Number/32; Int J = Number % 32; Bitset [I] | = (1 <j ); } Bool get (INT number) { Int I = Number/32; Int J = Number % 32; Return (bitset [I] & (1 <j ))! = 0; } Void output () { For (INT I = 0; I <m_nrange; I ++) { If (get (I )) { Cout <I <""; } } Cout <Endl; } PRIVATE: Int * bitset; Int m_nrange; // range of numbers Int m_nlength; // Len of Array }; Void print (int * a, int N) { For (INT I = 0; I <n; I ++) { Cout <A [I] <""; } Cout <Endl; } Void printduplicatenumbers (int * a, int N, int count) { Cout <"array numbers =============================\ N "; Print (A, N ); Cout <"duplicate numbers =============================\ N "; Bitset BS (count ); For (INT I = 0; I <n; I ++) { If (BS. Get (A [I]) Cout <A [I] <""; Else BS. Set (A [I]); } Cout <Endl; Cout <"existing numbers ===========================\ N "; BS. Output (); } Void test_defualt () { Const int n = 20; Const int range = 12; Srand (unsigned INT) Time (null )); Int A [n]; For (INT I = 0; I <n; I ++) { A [I] = rand () % range; } Printduplicatenumbers (A, N, range ); } Int _ tmain (INT argc, _ tchar * argv []) { Test_defualt (); Return 0; } /* Array numbers ================================== 7 0 2 8 0 3 0 3 2 1 7 5 11 5 4 11 1 0 2 4 Duplicate numbers ============================= 0 0 3 2 7 5 11 1 0 2 4 Existing numbers ================================ 0 1 2 3 4 5 7 8 11 */

[Link to this article]

Http://www.cnblogs.com/hellogiser/p/using-bitset-to-print-duplicate-elements-of-array.html

74 use bitset to output repeated elements in the array