求一组字符串最长前缀

package com.lkr.dataStructure;

import java.util.Scanner;

//求一组字符串最大前缀
public class ZuiChangSubStr {

    public static void main(String[] args){

        /*System.out.println("请输入一组字符串");
        Scanner input = new Scanner(System.in);
        String [] arrays = new String[5];
        for(int i = 0;i<arrays.length;i++){

            arrays[i] = input.next();
        }*/
        //String [] arrays = new String[]{""};
        //String [] arrays = new String[]{"abcd","ab","a"};
        //String [] arrays = new String[]{"abcd","","a"};
        String [] arrays = new String[]{"abcd","cf","a"};

        String result = longSubStr(arrays);
        if(result == ""){
            System.out.println("存在空字符串或者数组为空");
        }
        
        else
        System.out.println("最长前缀为："+result);

    }

    //求最长前缀
    public  static String longSubStr(String [] strings){

        String temp = strings[0];
        //数组为0情况或者第一个字符串为空
        if(strings.length == 0 || strings[0].equals("")){
            System.out.println("数组为空");
            return "";
        }
        //数组只有一个字符串情况
        if(strings.length == 1){
            return temp;
        }

        //遍历其他字符串进行比较
        for (int i = 1;i<strings.length;i++){
            //如果有其中一个字符串为空，则返回空
            if(strings[i].length() == 0){
                System.out.println("存在空字符串");
                return "";
            }
            //temp的字符串长度最小
            else if(temp.length() < strings[i].length()){
                for (int j=0;j<temp.length();j++){
                    if(temp.charAt(j) != strings[i].charAt(j)){
                        temp = strings[i].substring(0,j);
                        
                        break;
                    }

                }
            }
            //string[i]长度最小
            else  {
                for (int k=0;k<strings[i].length();k++){
                    if(temp.charAt(k) != strings[i].charAt(k)){
                        temp = strings[i].substring(0,k);
                        break;
                    }
                    //如果遍历完其中一个字符串，还没结束，返回该字符串
                    if(k == strings[i].length()-1){
                        temp = strings[i];
                    }
                }
            }
        }
        return temp;

    }
}