Try this:
In this example i use an array of chars, but you can substitute it with your integer array.
Weight list contains for each char the associated probability.
It represent the probability distribution of my charset.
In weightsum list for each char i stored his actual probability plus the sum of any antecedent probability.
For example in weightsum the third element corresponding to 'C', is 65:
P('A') + P('B) + P('C') = P(X=>c)
10 + 20 + 25 = 65
So weightsum represent the cumulative distribution of my charset.
weightsum contains the following values:
It's easy to see that the 8th element correspondig to H, have a larger gap (80 of course like his probability) then is more like to happen!
List<Character> charset = Arrays.asList('A','B','C','D','E','F','G','H','I','J');
List<Integer> weight = Arrays.asList(10,30,25,60,20,70,10,80,20,30);
List<Integer> weightsum = new ArrayList<>();
int i=0,j=0,k=0;
Random Rnd = new Random();
weightsum.add(weight.get(0));
for (i = 1; i < 10; i++)
weightsum.add(weightsum.get(i-1) + weight.get(i));
Then i use a cycle to get 30 random char extractions from charset,each one drawned accordingly to the cumulative probability.
In k i stored a random number from 0 to the max value allocated in weightsum.
Then i look up in weightsum for a number grather than k, the position of the number in weightsum correspond to the same position of the char in charset.
for (j = 0; j < 30; j++)
{
Random r = new Random();
k = r.nextInt(weightsum.get(weightsum.size()-1));
for (i = 0; k > weightsum.get(i); i++) ;
System.out.print(charset.get(i));
}
The code give out that sequence of char:
HHFAIIDFBDDDHFICJHACCDFJBGBHHB
Let's do the math!
A = 2
B = 4
C = 3
D = 5
E = 0
F = 4
G = 1
H = 6
I = 3
J = 2
Total.:30
As we wish D and H are have more occurances (70% and 80% prob.)
Otherwinse E didn't come out at all. (10% prob.)
