# Java algorithms: 1. skip steps; 2. Calculate the number of 1 in binary; 3. Calculate the number of 1 in decimal.

Source: Internet
Author: User

Some small algorithms are of the Java version, and a large number of questions on the network are aimed at C ++. Therefore, Java implementation is rare, but they are the basis of the test,

/**
* 1 hop steps
* Question: there are N levels in a step. If you can skip 1 level at a time, you can also skip 2 levels.
* Calculate the total number of hops in total and analyze the time complexity of the algorithm.
* It is a combination question, which is completely correct and difficult,
* The total number of M steps N is 2.
* O (m-N) * (2n-1 )))
*/
Public class jumpfootstep {

Public static void main (string [] ARGs ){
Jumpstepway (10 );
}

Private Static void jumpstepway (int n ){
Int twostep = 1;
While (2 * twostep <= N ){
Printleftjoin (twostep, N );
For (INT I = 1; I <= twostep; I ++ ){
Int onestep = N-2 * twostep;
List <integer> ls = new arraylist <integer> ();

For (Int J = 1; j <= onestep; j ++ ){
For (int K = 0; k <(twostep-I); k ++ ){
}
For (INT m = 0; m <j; m ++ ){
}
For (INT m = 0; m <I; m ++ ){
}
For (INT m = 0; m <onestep-J; m ++ ){
}
Print (LS );
Ls. Clear ();
}
If (I! = Twostep ){
For (Int J = 1; j <= onestep; j ++ ){

For (INT m = 0; m <j; m ++ ){
}
For (INT m = 0; m <I; m ++ ){
}
For (INT m = 0; m <onestep-J; m ++ ){
}
For (int K = 0; k <(twostep-I); k ++ ){
}

Print (LS );
Ls. Clear ();
}
}

}
Twostep ++;
}
}
Private Static void printleftjoin (INT twostep, int N ){
For (INT I = 1; I <= twostep; I ++ ){
System. Out. Print (2 );
}
For (INT I = 1; I <= N-2 * twostep; I ++ ){
System. Out. Print (1 );
}
System. Out. println ();
}
Private Static void print (list <integer> ls ){
For (INT m = 0; m <ls. Size (); m ++ ){
System. Out. Print (LS. Get (m ));
}
System. Out. println ();
}
}

There are repeated outputs in the answer to the step, which is a question to be resolved!

/**
* The binary value of 2 integers indicates the number of values in 1.
* Question: enter an integer to calculate the number of 1 in the Binary Expression of this integer.
* For example, input 10. Because the binary value is 1010 and there are two 1 s, Output 2.
*
* @ Author wangjichen
*
*/
Public class binaryonecount {

Public static void main (string [] ARGs ){

Long B = 1101;
Long I = 1;
Int COUNT = 0;
Int time = 0;
While (I> 0 ){
Long F = I & B;
F >>=time;
If (F = 1 ){
Count ++;
}
I <= 1;
Time ++;
}
System. Out. println (count );
}
}

Do not move the value of B to the right. If it is compared with I = 1, an endless loop occurs. Think about why...

/**
* 3 number of occurrences of 1 in a positive number from 1 to n
* Question: enter an integer n to calculate the number of times 1 appears in the decimal representation of the N integers from 1 to n.
* For example, input 12. integers from 1 to 12 contain numbers 1, 10, 11, and 12, and 1 appear 5 times in total. Analysis: This is a widely used Google interview question.
*
* @ Author wangjichen
*
*/
Public class decimalonecount {

Public static void main (string [] ARGs ){
Judge (111 );
}

Private Static void judge (int n ){
Int result = 0;
For (INT I = N; I> = 0; I --){
Result + = eachjudge (I );
}
System. Out. println (result );
}
Private Static int eachjudge (int n ){
Int COUNT = 0;
While (n> 0 ){
Int F = n % 10;
If (F = 1 ){
Count ++;
}
N/= 10;
}
Return count;
}
}

The answer to the second question is that when B is a negative number, it is an endless loop.

I don't know if the examiner wants to take this test, but the general idea is that the moving between the left and right is the dividing line of recruitment.

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