使用Linq Sum的Your way效率极低！
你可以通过直接跟踪和来大大改进这个问题，大约是原来的3倍，而不必总是一遍又一遍地迭代整个数组，只需要精确地跟踪你在数组中添加或删除的数量：

public void SplitWinnings()
{
    // Keep track of the sum yourself!
    // You already know how much you add each time!
    var sum = 0f;

    splitAmounts = new float[loseThreshold - 1];
    while(sum < totalWinAmount)
    {
        int rng = Random.Range(0, splitAmounts.Length);
        splitAmounts[rng] += minIncrement;
        // also keep track of the sum
        sum += minIncrement;

        float splitFloat = (Mathf.Round(splitAmounts[rng] * 100f) / 100f);

        if (splitFloat > 0)
        {
            // remove the previous value
            sum -= splitAmounts[rng];

            splitAmounts[rng] = splitFloat;

            // track the new value
            sum += splitFloat;
        }
    }
}

但是，您仍然要进行大量的舍入，并且可能以无效的迭代结束，这意味着放弃性能。
我甚至会更进一步，简单地预先计算0.05增量的最大倍数，迭代一个固定的数量-〉你确切地知道你需要预先填充多少元素，然后在int中完成所有剩下的计算。
这样做，我提高了100倍的时间！

public void SplitWinnings()
{
    // calculate how many of your minIncrements you need in total
    var totalMultiples = Mathf.RoundToInt(totalWinAmount / minIncrement);
    // while iterating will keep track of how many multiples you have remaining
    var remainingMultiples = totalMultiples;

    // Not sure why but you seem to require -1 here
    var splitCount = loseThreshold - 1;
    splitAmounts = new float[splitCount];
    
    // instead of while have a fixed amount of iterations
    // namely exactly for all return elements except the last one
    for(var i = 0; i < splitCount - 1; i++)
    { 
        // reducing the remaining array elements => save up at least 1 multiple for each array element
        var maxMultiples = remainingMultiples - (splitCount - i);
        // use at least one multiple per element
        // have to add +1 again since upper bound for int is exclusive but we want to use it up
        var multiples = Random.Range(1, maxMultiples + 1);
        
        // after calculating the multiples in int multiply again by the minIncrement to get according float value
        splitAmounts[i] = multiples * minIncrement;
        
        // and reduce the remaining multiples accordingly
        remainingMultiples -= multiples;
    }
    
    // The last item is just filled with whatever is left
    splitAmounts[splitCount - 1] = remainingMultiples * minIncrement;
}

参见.Net Fiddle进行比较

正如@Jeroen Mostert所提到的，这适用于较低数量的拆分值。但实际上正是我所寻找的。：）

public void SplitWinnings()
{
    splitAmounts = new float[loseThreshold - 1];
    while(splitAmounts.Sum() < totalWinAmount)
    {
        int rng = Random.Range(0, splitAmounts.Length);
        splitAmounts[rng] += minIncrement;

        float splitFloat = (Mathf.Round(splitAmounts[rng] * 100f) / 100f);

        if (splitFloat > 0)
        {
            splitAmounts[rng] = splitFloat;
        }
    }
}