How to choose 10 different integers from range (0, 99)

I want to choose 10 random integers from 0 开发者_如何学Pythonto 99. I know I can use:

random.randint(a, b)

But how to tell the randint() that I only want different integers.

Do I have to just check after each random generation to see if the integer has already been generated and call the method again? That does not seem like an optimal solution.

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from random import sample

sample(range(0, 100), 10)

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Here's general strategy that is language independent. Generate an array of 100 entries from 0 to 99. Choose a random number from 0 to 99 and swap the entry at that position with the element at position 0. Then successively choose a random number from i to 99, where i = 1 to 9 and swap the element at that position with the one at element i. Your 10 random numbers are in the first 10 positions of the array.

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This may not be a solution depending on what you will need to use this for... but have you thought of splitting it off into 10 different random number generators? Ex. 0-9, 10-19, 20-29, etc. I guess that's not really as "random" as you are specifying different ranges and are guaranteed 1 number from each range. The only other solution I can think of would be to have a list of the random integers, iterate through and check to see if the random number has been generated yet and if so run the random.randint() again.

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Here is a language independent solution:

integer numbers[10];
for(integer i = 0; i < 10; i += 1) {
    integer num = randomInteger(min = 0, max = (99 - i));
    boolean hasFoundDuplicate = false;
    for(integer j = 0; j < i && hasFoundDuplicate == false; j += 1) {
        if(numbers[j] == num) {
            num = 99 + 1 - i + j;
            hasFoundDuplicate = true;
        }
    }
    numbers[i] = num;
}