Leetcode-470- implement rand10() with rand7()

subject

Ideas

1. I have done this problem before , Although I have completely forgotten , I still can't figure it out , Go and see the solution .
2. For example, using rand2 seek rand4, If you use (rand2()-1+rand2()), It can be calculated that the range is [1,3] Between , Although the generated values are indeed [1,4] Between , But it doesn't meet the conditions .
3. But if it is 2*(rand2()-1)+rand2(), It can be calculated that [1,4] Between .
4. So obviously , Is to find out [1,7] turn [1,10] The formula of , Based on the above experience , In short, we must first turn out [0,x] come out , consider rand7()-1, Become [0,6] Value , In order to make its scope in 7 In multiple of , Put it x7, Become [0,42], To make the left boundary 7, Plus rand7(), Become [1,49]
5. here , have access to rand7() obtain rand49(), And each probability is uniform , namely 12345678910 These ten figures are uniform , Although the sum is not 1.
6. Then the follow-up is obvious , Abandon all not for [1,10] Or turn all you get into [1,10] that will do .

Code

    public int rand10() {
    
        while(true){
    
            // use rand7() Realization rand10()
            int num=0;
            num=(rand7() - 1) * 7 + rand7(); // Equal probability generation [1,49]
            if(num>=40) return num%10+1;
        }
    }