Question

我有一个练习，我需要找到最远和最接近的数字到一定数量（平均值）。有人能帮助我吗？这是我的代码：

yourclass.sigleton.pazienti()

Answer 1

如果您可以使用List<int>代替数组，那么根据您的要求，您可以更轻松地进行编码，也可能更清晰。

让我们改变一下：

int[] array = new int[10] {g1,g2,g3,g4,g5,g6,g7,g8,g9,g10};

进入这样的列表：

List<int> values = new List<int>(){g1,g2,g3,g4,g5,g6,g7,g8,g9,g10};

聚合将测试每个元素，直到列表完全经过测试。因此，我们可以尝试获得最接近和最远的值，如此

// value you want to check is the average of the list.
// Average is already implemented for the List
var value = values.Average();

// will keep the closest value in the collection to our value we are looking for
// and keep testing with the following value to end up with the final closest
var closestTo = values.Aggregate((x, y) => Math.Abs(x - value) < Math.Abs(y - value) ? x : y);

// same logic as closest except we keep the furthest
var furthestTo = values.Aggregate((x, y) => Math.Abs(x - value) > Math.Abs(y - value) ? x : y);

Answer 2

我希望这个例子可以帮助你，即使它是在java中。

    //Your array with the numbers
    int[] data = {12, 123, 65, 23, 4, 213, 32, 342, 21, 15};

    int testNum = 27; //Your test value

    int holderHigh = testNum;
    for(int i = 0; i < data.length; i++){//Goes through the array 
        //If the holder is lower than a value in the array it will get stored
        if(holderHigh < data[i]){ 
            holderHigh = data[i];
        }
    }

    int holderLow = testNum;
    for(int k = 0; k < data.length; k++){//Goes through the array 
        //If the holder is higher than a value in the array it will get stored  
        if(holderLow > data[k]){
            holderLow = data[k];
        }
    }        
    System.out.println(holderHigh + " and " + holderLow +
            " is the number farthest away from " + testNum);stNum);

找到最远和最接近的数字到数组c＃中的数字

2 个答案: