How to find the difference of two numbers and the absolute value of that answer without using math.abs function JAVA
A bit more formatted code.
import java.util.Scanner;
class A {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
double a;
double b;
System.out.println("Enter a: ");
a = in.nextDouble();
System.out.println("Enter b: ");
b = in.nextDouble();
double value = a - b;
//If value is negative...make it a positive number.
value = (value < 0) ? -value : value;
System.out.println("|"+a + "-" + b +"|" + " =" + value); // value should be printed here instead of (a-b) or (b-a)
System.out.println("|"+b + "-" + a +"|" + " =" + value);
}
}
Finding absolute value of a number without using Math.abs()
If you look inside Math.abs you can probably find the best answer:
Eg, for floats:
/*
* Returns the absolute value of a {@code float} value.
* If the argument is not negative, the argument is returned.
* If the argument is negative, the negation of the argument is returned.
* Special cases:
* <ul><li>If the argument is positive zero or negative zero, the
* result is positive zero.
* <li>If the argument is infinite, the result is positive infinity.
* <li>If the argument is NaN, the result is NaN.</ul>
* In other words, the result is the same as the value of the expression:
* <p>{@code Float.intBitsToFloat(0x7fffffff & Float.floatToIntBits(a))}
*
* @param a the argument whose absolute value is to be determined
* @return the absolute value of the argument.
*/
public static float abs(float a) {
return (a <= 0.0F) ? 0.0F - a : a;
}
How do I get the absolute value of an integer without using Math.abs?
You can use the conditional operator and the unary negation operator:
function absVal(integer) {
return integer < 0 ? -integer : integer;
}
get absolute value without using abs function nor if statement
From Bit Twiddling Hacks:
int v; // we want to find the absolute value of v
unsigned int r; // the result goes here
int const mask = v >> sizeof(int) * CHAR_BIT - 1;
r = (v + mask) ^ mask;
Find the minimal absolute value of a sum of two elements
For input arrays e.g({-1, -2, -3}
, {-1, -2}
, {-1}
your algorithm throws ArrayIndexOutOfBoundsException, so arrays when there are only negative numbers and there is no repeats
There is no chance to reach endless loop because either i or j change only + or - 1
How to get the difference between two integers
The result is 1 because compareTo()
returns 0 if the arguments are equal, -1 if the first int is smaller than the second one and 1 if the second one is smaller (you can read more about it in the official docs).
--> You should not use this method for this purpose. Calculate the difference instead:
int x = pos2.x - pos1.x;
int y = pos2.y - pos1.y;
int size = Math.abs(x * y);
Max absolute difference of two max values at the different parts of the array?
In your program:
leftMax[k] holds the greatest value in A[0],...,A[k].
rightMax[k] holds the greatest value in A[k],...,A[n-1].
However, the right part should start at index k+1, not at k.
Therefore I suggest you change this part:
for (int k = 0; k<A.length; k++){
dif = Math.abs(leftMax[k] - rightMax[k]);
if (dif>maxDif) {
maxDif = dif;
}
}
to
for (int k = 0; k<A.length - 1; k++){
dif = Math.abs(leftMax[k] - rightMax[k + 1]);
if (dif>maxDif) {
maxDif = dif;
}
}
In other words, the requirement is to compute:
Math.Abs(Max(A[0], A[1]...A[K]) - Max(A[K+1], A[K+2]... A[N-1]))
but I believe your current program computes:
Math.Abs(Max(A[0], A[1]...A[K]) - Max(A[k], A[K+1], A[K+2]... A[N-1]))
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