需要对字符串值进行排序，同时忽略对空字符串值的任何位置更改

Question

我正试图在Java中实现这样的场景。

我有一个ArrayList

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

目前，该列表是没有排序，我想它的方式，空字符串""的位置保持排序。

这意味着空字符串不排序，其他值是。 实施例 的ArrayList

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

注意，空字符串（最初在位置2,3 & 6,7,8）的位置被保持之后和分类仅与非空值来实现。我使用Java，我真的很想有一些想法来开始实现这个要求。我曾尝试谷歌，但无法找到一个开始就此。

请帮忙，

在此先感谢。

Answer 1

也许类似的东西：（贷汤姆他有用的链接）

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

也许你会想用这样的：

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)); 
    } 
} 

它运行在O（nlogn ）时间，因为排序，但这是可行的。