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How to compare floating point values with zero, I may get integers like 0,1,2 or floating point numbers 0.0 ,41.2 etc. How to compare these values against 0
I tried this way
String x="<some value either 0 or 1.2"
If(Integer.parseInt(x)==0)
System.out.println("parsed");
with input 0 it is working, with 1.2 throwing format exception.
Regards,
Raj
If your values can be decimal numbers, use this for parsing the string:
Double.parseDouble(x)
The comparison remains the same, though:
if (Double.parseDouble(x) == 0)
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I am trying to produce all permutations of an array of size N (i.e. N = 9) with possible elements Z (i.e. Z = [0,1,2,3,4])
Duplicates are allowed, but there needs to be at least one. I am able to write the algorithm to go through all possibilities but the minimum of at least one of each of Z is the part I'm having trouble with.
Looking to do this algorithm in java. Can someone help?
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I have a list of strings in random format:
AppName-ver-1.1.0-data.exe
AppName-ver-1.1.1-secondData.exe
AppName-ver-1.2.0-data.exe
AppName-ver-1.2.1-data.exe
AppName-ver-1.2.3-data.exe
AnotherAppName-ver-1.0.0-data.exe
AnotherAppName-ver-1.0.0-secondData.exe
What would be an efficient way in java to find the closest value to string:
AppName-ver-1.2.4-data.exe
UPD: closest - by the naming not length so AppName-ver-1.2.3-data.exe is the expected result
To emphasize what commenters already pointed out:
If you define 'closest' to be the string length, then
AppName-ver-1.2.4-data.exe has the value 26, and
AppName-ver-1.1.0-data.exe
AppName-ver-1.2.0-data.exe
AppName-ver-1.2.1-data.exe
AppName-ver-1.2.3-data.exe
all resemble 26 as well so they are a direct match.
You could also define 'closest' to have the least Hamming Distance. This will give completely different results and AppName-ver-1.2.0-data.exe might win as it is just one bit off.
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I tried it with Sieve Of Eratosthenes, but I quickly run into the problem that my boolean array can't go past Integer.MAX
How should I approach this problem?
Arrays
Use 2 or more dimensional maps. So converting long to two integers (for accessing the arrays) will be like this:
array[N / Integer.MAX_VALUE][N % Integer.MAX_VALUE] where N is long and array is the boolean array.
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I'm trying to find out an algorithm that returns the number that has more occurrences in an array that contains number from 0 to 9 and that has a complexity of n.
I though to using an HashMap but it would require n^2
If anyone can write the code down,i'd prefer in Java but pseudocode is the same
Use ten counters (one per digit), scan the array and increment the counters corresponding to the digits. (You are actually computing the histogram of the digit frequencies.)
Report the digit with the largest counter.
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I want to know where is the difference between System.out.write and System.out.print in my Java class.
The two methods of PrintStream have different meanings:
print(int) writes a decimal representation of the entire int, while
write(int) writes the least significant byte of the specified int to the output.
This leads to different results: if you call print(48), the output is going to be 48, but if you call write(48), the output would be system-dependent, but on most systems it would be 0.
Demo.