产生范围内唯一的编号（0 - X），保持历史prevent重复 [英] Generate unique number within range (0 - X), keeping a history to prevent duplicates

问题描述

我遇到了，我需要从返回给定范围内的随机数的函数挑战0 - X 。不仅如此，但我需要返回的数字为唯一;没有复制一个已经在previous调用该函数返回的数字。

I ran into the challenge where I need a function that returns a random number within a given range from 0 - X. Not only that, but I require the number returned to be unique; not duplicating numbers that have already been returned on previous calls to the function.

任选地，当这样做（例如，范围已被用尽），只是范围内返回一个随机数。

Optionally, when this is done (e.g. the range has been 'exhausted'), just return a random number within the range.

如何将一个去这样做呢？

How would one go about doing this?

推荐答案

我写了这个功能。它保持了自己的阵列生成的数字，preventing初始重复的历史，继续输出一个随机数，如果在范围内的所有数字已输出一次：

I wrote this function. It keeps its own array with a history of generated numbers, preventing initial duplicates, continuing to output a random number if all numbers in the range have been outputted once:

// Generates a unique number from a range
// keeps track of generated numbers in a history array
// if all numbers in the range have been returned once, keep outputting random numbers within the range
var UniqueRandom = { NumHistory: new Array(), generate: function(maxNum) {
        var current = Math.round(Math.random()*(maxNum-1));
        if (maxNum > 1 && this.NumHistory.length > 0) {
            if (this.NumHistory.length != maxNum) {
                while($.inArray(current, this.NumHistory) != -1) { current = Math.round(Math.random()*(maxNum-1)); }
                this.NumHistory.push(current);
                return current;
            } else {
                //unique numbers done, continue outputting random numbers, or we could reset the history array (NumHistory = [];)
                return current;
            }
        } else {
            //first time only
            this.NumHistory.push(current);
            return current;
        }
    }
};

这里的工作小提琴

我希望这是使用的人！

编辑：由尖尖指出下面时，它可能会以缓慢的大范围（这里是一个
小提琴，会在一个范围0-1000 ，这似乎运行正常）。然而;我并不需要一个非常大的范围内，所以也许这个功能确实不适合，如果你看看生成和保持一个巨大的范围内跟踪。

as pointed out by Pointy below, it might get slow with a large range (here is a fiddle, going over a range from 0-1000, which seems to run fine). However; I didn't require a very large range, so perhaps this function is indeed not suited if you look to generate and keep track of an enormous range.

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