[Link to this article]
Http://www.cnblogs.com/hellogiser/p/findnnumberswhichappearonlyonceinarray.html
【Question]
Three numbers a, B, and c appear only once in an array, and the other numbers appear twice. Find three numbers that appear only once.
【Analysis]
This is a novelBitwise operation. Previous blog posts34. two numbers that appear only once in the array [Find two numbers which appear once]Analysis of N = 1 and N = 2.
(1). N = 1, the result of all numbers in the array is.
(2 ). when N = 2, all numbers in the array are exclusive or the result is equal to a ^ B. The array is divided into two groups based on the position where the last 1 in the binary of a ^ B; returns a and B if the numbers in each group are different or different.
(3) When. N = 3, all numbers in the array are exclusive or the result is equal to a ^ B ^ c. How can we differentiate them? If we can find a number that appears only once, the remaining two digits that appear only once can be converted to N = 2.
The specific ideas are as follows:
(1). f (x) = x & (x) The result is the position of the last 1 of x.
(2). x = a ^ B ^ c.
(3 ). flag = f (x ^ a) ^ f (x ^ B) ^ f (x ^ c) the result must be 1, because f (m) ^ f (n) the result is 0 or 2.
(4 ). f (x ^ a) ^ f (x ^ B) ^ f (x ^ c) where the mbit is 1, then x ^ a, x ^ B, x ^ c must have 1 or 3 mth digits as 1.
(5). We can use the reverse identification method to obtain that x ^ a, x ^ B, and x ^ c have only one mbit as 1.
For example, data = {1, 2, 4, 4, 5, 5, 6}
X = a ^ B ^ c = 1 ^ 2 ^ 3 = 000
X ^ a = 001, x ^ B = 010, x ^ c = 011
F (x ^ a) = 001, f (x ^ B) = 010, f (x ^ c) = 001
Flag = f (x ^ a) ^ f (x ^ B) ^ f (x ^ c) = 010, flag = f (flag) = 010
F (x ^ B) = flag
First = B = 1
Second = 3, third = 1
The complete code is as follows:
【Code]
C ++ Code
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// 58_FindNumbersAppearOnce.cpp: Defines the entry point for the console application. // /* Version: 1.0 Author: hellogiser Blog: http://www.cnblogs.com/hellogiser Date: 2014/5/27 */
# Include "stdafx. h"
// Find number which appear once Void Find1NumberAppearOnce (int data [], int length, int & num) { If (NULL = data  length <1) Return;
// Get the exclusive or result of array // Int xor = 0; For (int I = 0; I <length; ++ I) Xor ^ = data [I]; Num = xor; }
// Get last 1 bit of n // N = 00001110 > 00000010 Unsigned int GetLast1Bit (int n) { Return n & (n ); }
// Find 2 numbers which appear once Void Find2NumbersAppearOnce (int data [], int length, int & num1, int & num2) { If (NULL = data  length <2) Return;
// Get the exclusive or result of array // A ^ B Int xor = 0; For (int I = 0; I <length; ++ I) Xor ^ = data [I];
// Find the last bit 1 of xor Int flag = GetLast1Bit (xor ); Num1 = num2 = 0; For (int j = 0; j <length; ++ j) { // Divide numbers in data into 2 groups by flag: // Numbers in group1: the & result is 1 // Numbers in group2: the & result is 0 If (flag & data [j]) { Num1 ^ = data [j]; } Else { Num2 ^ = data [j]; } } }
// Swap a and B Void myswap (int & a, int & B) { Int t =; A = B; B = t; }
// Find 3 numbers which appear once /* (1). f (x) = x & (x) (2). x = a ^ B ^ c (3). flag = f (x ^ a) ^ f (x ^ B) ^ f (x ^ c) (4). flag = f (flag) (5). x ^ a, x ^ B, x ^ c, only one of three is 1 at mbit F (x ^ a) = flag
For example: Data = {1, 2, 3, 4, 5, 5, 6} X = a ^ B ^ c = 1 ^ 2 ^ 3 = 000 X ^ a = 001, x ^ B = 010, x ^ c = 011 F (x ^ a) = 001, f (x ^ B) = 010, f (x ^ c) = 001 Flag = f (x ^ a) ^ f (x ^ B) ^ f (x ^ c) = 010 Flag = f (flag) = 010 F (x ^ B) = flag First = B = 1 Second = 3, third = 1 */ Void Find3NumbersAppearOnce (int data [], int length, int & num1, int & num2, int & num3) { If (NULL = data  length <3) Return;
// Get the exclusive or result of array // A ^ B ^ c Int xor = 0; For (int I = 0; I <length; I ++) Xor ^ = data [I];
Int flag = 0; For (int I = 0; I <length; I ++) Flag ^ = GetLast1Bit (xor ^ data [I]); Flag = GetLast1Bit (flag );
// Get the first unique number Int first = 0; For (int I = 0; I <length; I ++) If (GetLast1Bit (data [I] ^ xor) = flag) First ^ = data [I]; Num1 = first;
// Move the first number to the end of array For (int I = 0; I <length; I ++) { If (first = data [I]) { Myswap (data [I], data [length1]); Break; } }
// Get the second and third unique number Find2NumbersAppearOnce (data, length1, num2, num3 ); }
// ================================================ ====================================== // Test cases Void test_base1 (int data [], int length) { Int num1; Find1NumberAppearOnce (data, length, num1 ); Printf ("% d \ n", num1 ); }
Void test_base2 (int data [], int length) { Int num1, num2; Find2NumbersAppearOnce (data, length, num1, num2 ); Printf ("% d \ n", num1, num2 ); }
Void test_base3 (int data [], int length) { Int num1, num2, num3; Find3NumbersAppearOnce (data, length, num1, num2, num3 ); Printf ("% d \ n", num1, num2, num3 ); }
Void test_case1 () { Int data [] = {1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6 }; Int length = sizeof (data)/sizeof (int ); Test_base1 (data, length ); }
Void test_case2 () { Int data [] = {1, 2, 3, 3, 4, 4, 5, 5, 6, 6 }; Int length = sizeof (data)/sizeof (int ); Test_base2 (data, length ); }
Void test_case3 () { Int data [] = {1, 2, 3, 4, 4, 5, 5, 6, 6 }; Int length = sizeof (data)/sizeof (int ); Test_base3 (data, length ); }
Void test_main () { Test_case1 (); // 1 Test_case2 (); // 1 2 Test_case3 (); // 2 3 1 } // ================================================ ======================================
Int _ tmain (int argc, _ TCHAR * argv []) { Test_main (); Return 0; } 
[Reference]
Http://www.cnblogs.com/hellogiser/p/3738909.html
Http://zhedahht.blog.163.com/blog/static/25411174201283084246412/
Http://www.cnblogs.com/youxin/p/3349834.html
Http://www.cnblogs.com/kedebug/archive/2012/12/22/2829013.html
[Link to this article]
Http://www.cnblogs.com/hellogiser/p/findnnumberswhichappearonlyonceinarray.html