[algorithm] bit operation two

Topic One:

Given an array of shaped arr and an integer greater than 1 K. It is known that only 1 numbers in arr appear odd times, and the other numbers appear even several times, so return the odd number of occurrences.

The time complexity is O (N) and the additional space complexity is O (1).

Ideas:

The result of an integer n with 0 xor is n, the result of an integer n with an integer n is 0. So first apply for an orthopedic variable, which is written as EO. With EO and each number XOR (the current number of eo=eo^), the last EO value is the number that has an odd number of occurrences.

An XOR operation satisfies the commutative law and the binding law.

 Public Static void printOddTimesNum1 (int[] arr) {        int eO = 0;          for (int  cur:arr) {            ^= cur;        }        System.out.println (EO);    }

Topic Two:

Given an array of shapes arr, there are two numbers that appear odd several times, and the other numbers appear even several times. Print these two numbers.

Ideas:

If only A and B appear odd number of times, then the final XOR result EO must be a^b. Therefore, if there are two occurrences of an odd number in the array, the final result EO must not be 0. Then it is certainly possible to find a bit bit that is not equal to 0 on the 32-bit integer eo, assuming that k is not equal to 0. EO is not equal to 0 on the K-bit, stating that K-bits A and B must have one 1 and the other is 0. Next, set a variable to Ehasone, and then iterate through the array again. In this traversal, the Ehasone only has an integer that is 1 on the K-bit and the other number is ignored. Then at the end of the second traversal, Ehasone is one of a or B, and Eo^ehasone is another number.

 Public Static voidPRINTODDTIMESNUM2 (int[] arr) {        intEO = 0, Eohasone = 0;  for(intCurnum:arr) {EO^=Curnum; }        intRightone = EO & (~eo + 1);  for(intCur:arr) {            if((cur & rightone)! = 0) {Eohasone^=cur; }} System.out.println (Eohasone+ "" + (EO ^eohasone)); }

Topic Three:

Given an array of shaped arr and an integer greater than 1 K. It is known that only 1 numbers in arr have occurred 1 times, and the other numbers have been k times, please return only 1 occurrences.

Ideas:

Two k-binary numbers A and B, the result of no carry addition on the I bit is (a (i) +b (i))%k. So if K is the same K-binary number without carrying the sum, the result must be that each one is 0 K-binary number.

First set up a variable eo, which is a 32-bit K-binary number, and each position is 0. then traverse arr to convert each integer that is traversed to a K-number, and then add the non-rounding to the EO. At the end of the traversal, converting the 32-bit K-eores into a decimal integer is the final result. Because K is the same K-binary number is not added, the result must be every one on the 0 K-number, so only one occurrence of the number will be left.

 Public Static intOncenum (int[] arr,intk) {int[] EO =New int[32];  for(inti = 0; I! = arr.length; i++) {Setexclusiveor (EO, Arr[i], k); }        intres =Getnumfromksysnum (EO, k); returnRes; }    //Adds the current number corresponding to the K-binary to the corresponding position of the array     Public Static voidSetexclusiveor (int[] EO,intValueintk) {int[] Curksysnum =Getksysnumfromnum (value, k);  for(inti = 0; I! = eo.length; i++) {Eo[i]= (Eo[i] + curksysnum[i])%K; }    }    //Convert value to 32-bit K-binary     Public Static int[] Getksysnumfromnum (intValueintk) {int[] res =New int[32]; intindex = 0;  while(Value! = 0) {Res[index+ +] = value%K; Value= value/K; }        returnRes; }    //get the final result based on the remaining digits in the last array     Public Static intGetnumfromksysnum (int[] EO,intk) {intres = 0;  for(inti = eo.length-1; I! =-1; i--) {res= Res * k +Eo[i]; }        returnRes; }

[algorithm] bit operation two