从随机选择器列表中排除数字

时间:2018-06-03 04:14:41

标签: java android game-development

我有一个随机选择器代码,可以从1到6中选择随机数。 你可以给我一个方法,如何从随机选择列表中排除挑选的数字..

import java.util.Random;

Random rand = new Random();

int  n = rand.nextInt(6) + 1;

就像前者一样: 1.2.3.4.5.6 随机选择= 5 1.2.3.4.6 随机选择= 2 1.3.4.6 等等 ty提前家伙

2 个答案:

答案 0 :(得分:0)

将所有有效数字放入ArrayList,然后从列表中选择一个随机索引。然后从列表中删除该号码并重复。

我的Java有点生疏,所以希望我写的代码是有道理的:

ArrayList<int> validOptions = /**/; // make your list with all initial options

int firstIndex = random.Next(validOptions.count());
int firstPick = validOptions.get(firstIndex);
validOptions.removeAt(firstIndex);

int secondIndex = random.Next(validOptions.count());
int secondPick = validOptions.get(secondIndex);
validOptions.removeAt(firstIndex);

答案 1 :(得分:0)

您可以将已经拾取的数字添加到ArrayList并选择一个数字,直到该数字未包含在列表中。

// list of numbers that I already picked
ArrayList<Integer> randomNumbersPicked = new ArrayList<>();
// int to save the current random number
int myCurrentRandomNumber;

while(iNeedAnotherNumber){
    do {
        myCurrentRandomNumber = generateRandomNumber(a, b);
    //repeat this until the number is not in the list
    } while (randomNumbersPicked.contains(new Integer(myCurrentRandomNumber)));
    //here there is a unique random number, do what you will
    System.out.println("A new number has been picked: " + myCurrentRandomNumber);
    //add the number to the list so it wont be picked again
    randomNumbersPicked.add(new Integer(myCurrentRandomNumber));
}

最诚挚的问候! Dknacht。