java - 如何对数组中的大量数字进行冒泡排序？

Question

我的问题是-

已提供字符串数组 nums 作为输入。每个字符串代表一个非负整数。整数可能非常大，因此已存储为字符串。该数组必须按字符串整数值的升序排序。 bubbleSort({"999","723534","99","18456543876","54253445340001","98","112343",})

预期输出为-{"98","99","999","112343","723534","18456543876","54253445340001"}

实际输出为-{"99","98","999","112343","723534","18456543876","54253445340001"}

我不知道如何比较相同数字的大数。

public class BubbleSortLargeNums {

    static String[] testcase1 = {"999","723534","99","18456543876","54253445340001","98","112343",};
    //static String[] testcase1 = {"1"};

    public static void main(String args[]){
        BubbleSortLargeNums testInstance = new BubbleSortLargeNums();
        String[] result = testInstance.bubbleSort(testcase1);
        System.out.print("{");
        for (int i=0;i<result.length;i++){
            if (i>0)
                System.out.print(",");
            System.out.print('"'+result[i]+'"');
        }
        System.out.println("}");
    }

    //write your code here
    public String[] bubbleSort(String[] arr){
        int j=0;
        int prevI=0;
        for(int i=0;i<arr.length-1;i++){
            j++;
            if(j<arr.length){
                i=prevI;
            }
            else{
                j=i+1;
            }
            if(arr[i].length()>arr[j].length()){
                String temp=arr[i];
                arr[i]=arr[j];
                arr[j]=temp;
            }
            prevI=i;
        }
        return arr;
    }
}

Answer 1

首先缩短元素的长度

if (arr[i].length()>arr[j].length()) { 
String temp=arr[i];
arr[i]=arr[j];
arr[j]=temp;
}

现在问题将出在相同长度的元素例如98 和 99。当结果 > 1 时使用 compareTo 运算符交换

if j > j+1 交换数字，例如"99".compareTo."98"=1

计算第一位数字的差异，但如果第一位数字相同，则比较下一位数字并计算差异(左--->右)

if(str[j].length() == (str[j+1]).length()){ 
if ( str[j].compareTo(str[j+1])>0 ){
String temp = str[j];               
str[j] = str[j+1];
str[j+1] = temp;
   }
}