当列表中不存在X时，从列表中查找大于X的值 [英] Find greater than X from list when X doesn't exist in list

问题描述

我正在尝试从列表中查找大于某个值(在我的情况下为已知)的值.

I am trying to find values from a list that are greater than certain value (known in my case).

示例:

给出

list = [1, 2, 5, 10, 15];  //list is sorted

查找大于X的值(在这种情况下为=7).

Find values greater than X (=7 in this case).

所需结果=返回值= [10, 15]

Desired result = Return a list with values = [10, 15]

我尝试使用Java二进制搜索，例如

I tried using java binary search, like

int index = Collections.binarySearch(list, X);

我的计划是找到索引(X)，然后返回索引之后的所有元素.

My plan was to find index (of X) and then return all the elements after index.

但是索引返回负值，据我了解，因为7不在列表中.

But index return negative, that I understand because 7 is not in the list.

还有其他方法吗?有人建议.

Is there any other way ? Someone please suggest.

推荐答案

如果您的列表比

If your list is sorted than Collection#binarySearch will return the index of the search key, if it is contained in the list; otherwise, (-(insertion point) - 1). You can calculate begin index of insertion_point like below:

     index= -(insertion_point) - 1
     -(insertion_point)= index+1
     insertion_point=-(index+1)

在获得List的开始索引之后，您可以应用

After got the begin index of List than you can apply subList method to got the resulted list greater than X.

    Integer[] values = {1, 2, 5, 10, 15};
    List<Integer> list = new ArrayList<>(Arrays.asList(values));
    int X = 7;
    int index = Collections.binarySearch(list, X);

    int insertion_point = -(index+1); // As calculated above.

    List<Integer> res = list.subList(insertion_point, list.size());
    System.out.println(res);

输出:[10，15]

这篇关于当列表中不存在X时，从列表中查找大于X的值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！