如何在 java 中不使用 sort() 方法对字符串进行排序？

Question

I want to sort a string "computer" -> "cemoprtu", but without using Arrays.sort(string).

Answer 1

Looks like you need to sort the characters, so I'd start with

String input = "computer";
char[] characters = input.toCharArray();
//now sort characters using some algorithm
String output = new String(sorted_characters); //sorted_characters might be characters after sorting, if you sort in place

Answer 2

package practice;


class Practice
{
public static void main(String args[])
{
   String s = "bcfaed";
   char a[]=s.toCharArray();
   char b[]=new char[a.length];
   int count=0;
   for(int i=0;i<a.length;i++)
   {
       count=0;
       for(int j=0;j<a.length;j++)
       {
       if(a[i]<a[j])
       {
       count++;
       }
       }
       b[count]=a[i];
   }
   for(char x:b)
   {
       System.out.println(x);
   }
}
}

Answer 3

查看不同的排序算法并实现一些，尝试冒泡然后快速排序。

Answer 4

class Abc { public static void main(String[] args) { String str = "welcome"; char temp = 0; char arr[] = str.toCharArray(); for (int i = 0; i < arr.length; i++) { for (int j = 0; j < arr.length; j++) { if (arr[j] > arr[i]) { temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } } } for (int i = 0; i < arr.length; i++) { System.out.print(arr[i]); } } }

Answer 5

class Abc {

public static void main(String[] args) {
    String str = "welcome";
    char temp = 0;

    char arr[] = str.toCharArray();
    for (int i = 0; i < arr.length; i++) {
        for (int j = i+1; j < arr.length; j++) {
            if (arr[j] > arr[i]) {
                temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }
    }

    for (int i = 0; i < arr.length; i++) {
        System.out.print(arr[i]);
    }
}

}