Question

我有这段代码需要根据给定的Double distance输入返回一个索引：

public static int calculateHashPrecision(Double reference) {
    double[][] ranges = {
        {0.037, 0.018},
        {0.149, 0.149},
        {1.19, 0.6},
        {4.47, 4.78},
        {38.2, 19.1},
        {152.8, 152.8},
        {1200, 610},
        {4900, 4900},
        {39000, 19500},
        {156000, 156000},
        {1251000, 625000},
        {5004000, 5004000},
    };
    return 6;
  }

假设传递的reference的值是1000，所以它落在1200, 610范围内，因此此函数应返回索引6。对此的最佳实现是什么功能？

更新：而且如果它不在特定的硬编码范围内，它将返回最接近的那个。

Answer 1

对于每个循环，您都可以更快，更快速地完成

    int index = 0;
    for(double[] range: ranges){
        if( range[1] < reference && reference < range[0]){
           return index;
        }
        index ++;
    }

Answer 2

public static int calculateHashPrecision(Double reference) {
double[][] ranges = {
    {0.037, 0.018},
    {0.149, 0.149},
    {1.19, 0.6},
    {4.47, 4.78},
    {38.2, 19.1},
    {152.8, 152.8},
    {1200, 610},
    {4900, 4900},
    {39000, 19500},
    {156000, 156000},
    {1251000, 625000},
    {5004000, 5004000},
};
int maxValueIndexInRaw = 0;
int minValueIndexInRaw = 1;
for (int i = 0; i < ranges.length; i++) {
    if (reference >=ranges[i][minValueIndexInRaw] && reference <= ranges[i][maxValueIndexInRaw] ) {
      return i;
    }
}
return -1;

}

Answer 3

public static int calculateHashPrecision(double reference) {
    double[][] ranges = {
        {0.037, 0.018},
        {0.149, 0.149},
        {1.19, 0.6},
        {4.47, 4.78},
        {38.2, 19.1},
        {152.8, 152.8},
        {1200, 610},
        {4900, 4900},
        {39000, 19500},
        {156000, 156000},
        {1251000, 625000},
        {5004000, 5004000},
    };
    // search through the array of ranges, and return the first index it falls into
    for (int i = 0; i < ranges.length; i++) {
        double[] rng = ranges[i];
        // ranges as given have the higher number first, lower number second
        // adjust this logic as necessary
        double rng_bottom = rng[1];
        double rng_top = rng[0];
        if (rng_bottom < reference && reference < rng_top) {
            return i;
        }
    }
    // if no range contains the reference, e.g. for the number 6000, then return -1
    return -1;
}

您可能需要仔细检查您的范围，以确保它们被正确编码。您是说一半的人都具有相同的开始和结束吗？

Answer 4

如果数据已排序，则可以使用二进制搜索

    int hi = ranges.length - 1, lo = 0;
    while (lo <= hi) {
      int mid = lo + (hi - lo) / 2;
      int cmp = Double.compare(ranges[mid][0], reference);
      if (cmp > 0 || cmp == 0) {
        if (ranges[mid][1] <= reference) return mid;
        else hi = mid - 1;
      } else if (cmp < 0) {
        lo = mid + 1;
      }
    }
    return -1;

计算给定输入是否落入给定值范围内

4 个答案: