33 把数组排成最小数

描述：









实现代码：

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;

public class Solution {
public static void main(String[] args) {
int arr[]={3,32,321};
int arr1[]={3,32,321};

System.out.println("方法一");
String res=PrintMinNumber(arr);
System.out.println(res);

// 方法二
String res2=PrintMinNumber2(arr1);
System.out.println("方法二");
System.out.println(res);

}
public static String PrintMinNumber2(int [] numbers) {
int n;
String s="";
ArrayList<Integer> list=new ArrayList<Integer>();
n=numbers.length;

for(int i=0;i<n;i++){
list.add(numbers[i]);//将数组放入arrayList中
}
//实现了Comparator接口的compare方法，将集合元素按照compare方法的规则进行排序
Collections.sort(list,new Comparator<Integer>(){

@Override
public int compare(Integer str1, Integer str2) {
// TODO Auto-generated method stub
String s1=str1+""+str2;
String s2=str2+""+str1;

return s1.compareTo(s2);
}
});

for(int j:list){
s+=j;
}
return s;
}
public static String PrintMinNumber(int [] numbers) {
String res="";
ArrayList<String> minNumber=new ArrayList<String>();
for (int i : numbers) {
minNumber.add(String.valueOf(i));
}
Collections.sort(minNumber, new Comparator<String>() {
@Override
public int compare(String a, String b) {
int len1 = a.length();
int len2 = b.length();
int lim = Math.min(len1, len2);
char v1[] = a.toCharArray();
char v2[] = b.toCharArray();

int k = 0;
while (k < lim) {
char c1 = v1[k];
char c2 = v2[k];
if (c1 != c2) {
return c1 - c2;
}
k++;
}

if (len1>lim) {
return compare(a.substring(lim), b);
}else if (len2>lim) {

return compare(a,b.substring(lim));
}else {
return len1 - len2;
}

}
});

//System.out.println(minNumber);
StringBuilder sb=new StringBuilder();
for (String i : minNumber) {
sb.append(i);
}
return sb.toString();
}
}

 测试结果：