Question

我正在尝试用Java实现这样的场景。

我有一个ArrayList

["23","5","","","54","20","","","","0"]`

目前，列表未排序，我想以保持空字符串""的位置的方式对其进行排序。

这意味着空字符串不会被排序，其他值也是如此。示例 ArrayList的

["0","5","","","20","23","","","","54"].

请注意，之后保留空字符串的位置（最初位于2,3和6,7,8位置），并且仅使用非空值进行排序。我正在使用Java，我真的想要一些想法来开始实现这个要求。我试过谷歌，但无法找到一个良好的开端。

请帮忙，

提前致谢。

Answer 1

也许是这样的:(致信 Tom 以获得有用的链接）

    String[] arr = new String[] { "23","5","","","54","20","","","","0" }; // the original array
    boolean[] space = new boolean[arr.length]; // indicates whenever the index contains an empty space
    String[] target = new String[arr.length]; // the reslted array

    for (int i = 0; i < space.length; i++) // init spaces
    {
        space[i] = arr[i].isEmpty();
    }

    Arrays.sort(arr); // sort the original array

    int index = 0;
    for (int i = 0; i < arr.length; i++)
    {
        if (arr[i].isEmpty()) continue; // just a space ignore that
        index = i; // real values start here
        break;
    }

    for (int i = 0; i < space.length; i++)
    {
        if (space[i] == true)
        {
            target[i] = ""; // restore space
        }
        else
        {
            target[i] = arr[index]; index++;
        }
    }

Answer 2

可能你会想要使用这样的东西：

import java.util.Arrays;
import java.util.LinkedList;
import java.util.TreeSet;


public class StringSort {

    public static void main(String[] args) {
        String[] numbers = new String[] {"23","5","","","54","20","","","","0"};
        // Save the indices to be able to put the sorted numbers back in the array
        LinkedList<Integer> indices = new LinkedList<>();
        // Keep an ordered list of parsed numbers
        TreeSet<Integer> set = new TreeSet<>();
        // Skip empty entries, add non-empty to the ordered set and the indices list
        for (int i = 0; i < numbers.length; i++) {
            if (numbers[i].equals(""))
                continue;
            set.add(Integer.parseInt(numbers[i]));
            indices.add(i);
        }

        // Put the ordered integers back into the array
        for (int i : set) {
            numbers[indices.pop()] = Integer.toString(i);
        }
        System.out.println(Arrays.toString(numbers));
    }
}

由于排序，它在O（nlogn）时间运行，但这是可行的。

需要对String值进行排序，同时忽略对空字符串值的任何位置更改

2 个答案: