Question

我有List的{{1}}：

Doubles

我需要从该列表中找到给定值的最接近值。我在下面的解决方案中尝试了SO的另一个答案，但是到目前为止，这仍然行不通，List<Double> aList = Arrays.asList(100.33,300.99,600.0,900.44,1200.88,1500.15);始终为0。

下面是我正在检查的解决方案：

idx

Answer 1

您应该使用差的绝对值，并且它可能大于或小于0

尝试

private Double chekcer() {
  int myNumber = 900;
  double distance = Math.abs(aList.get(0)- myNumber);
  int idx = 0;
  for (int c = 1; c < aList.size(); c++) {
    double cdistance = Math.abs(aList.get(c) - myNumber);
    if (cdistance < distance) {
        idx = c;
        distance = cdistance;
    }
  }
  Log("",...);
  return aList.get(idx);
}

修改

这是我的全部代码，

public static void main(String[] args)  {

    System.out.println(chekcer());
}

private static Double chekcer() {
      List<Double>  aList = Arrays.asList(100.33,300.99,600.0,900.44,1200.88,1500.15);
      int myNumber = 900;
      double distance = Math.abs(aList.get(0)- myNumber);
      int idx = 0;
      for (int c = 1; c < aList.size(); c++) {
        double cdistance = Math.abs(aList.get(c) - myNumber);
        if (cdistance < distance) {
            idx = c;
            distance = cdistance;
        }
      }
      return aList.get(idx);
    }

输出

900.44

Answer 2

在Java 8中：您可能会发现该代码很有用

    private static Double chekcer() {
      List<Double>  aList = Arrays.asList(100.33,300.99,600.0,900.44,1200.88,1500.15);
      int myNumber = 900;
      return aList.stream().min(Comparator.comparingDouble(i -> Math.abs(i - myNumber))).orElse(null);     
     }

从双数列表中获取给定值的最接近索引值

2 个答案: