蓝桥杯java 基础训练 十六进制转八进制

问题描述
　　给定n个十六进制正整数，输出它们对应的八进制数。

输入格式
　　输入的第一行为一个正整数n （1<=n<=10）。
　　接下来n行，每行一个由0~9、大写字母A~F组成的字符串，表示要转换的十六进制正整数，每个十六进制数长度不超过100000。

输出格式
　　输出n行，每行为输入对应的八进制正整数。

　　【注意】
　　输入的十六进制数不会有前导0，比如012A。
　　输出的八进制数也不能有前导0。

样例输入
　　2
　　39
　　123ABC

样例输出
　　71
　　4435274

一、

import java.util.*;
public class Main{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        String[] sts = new String[n];
        for(int i=0;i<n;i++){
            sts[i] = sc.next();
        }
        sc.close();
        for(int i=0;i<n;i++){
            String strBinary = toBinary(sts[i]);
            int len_strBin = strBinary.length();
            if(len_strBin%3==1) strBinary = "00"+strBinary;
            if(len_strBin%3==2) strBinary = "0"+strBinary;
            String strOctal = toOctal(strBinary);
            System.out.println(strOctal);
        }
    }
    private static String toOctal(String strBinary) {
        int len = strBinary.length();
        int k;
        StringBuffer stb = new StringBuffer();
        if(strBinary.substring(0, 3).equals("000"))
            k=3;
        else
            k=0;
        for(int i=k;i<len-2;i+=3){
            switch (strBinary.substring(i, i+3)) {
                case "000":stb.append("0");break;
                case "001":stb.append("1");break;
                case "010":stb.append("2");break;
                case "011":stb.append("3");break;
                case "100":stb.append("4");break;
                case "101":stb.append("5");break;
                case "110":stb.append("6");break;
                case "111":stb.append("7");break;
                default:break;
            }
        }
        return stb.toString();
    }
    private static String toBinary(String strHex) {
        int len_str = strHex.length();
        StringBuffer stb = new StringBuffer();
        for(int i=0;i<len_str;i++){
            switch (strHex.charAt(i)) {
                case '0':stb.append("0000");break;
                case '1':stb.append("0001");break;
                case '2':stb.append("0010");break;
                case '3':stb.append("0011");break;
                case '4':stb.append("0100");break;
                case '5':stb.append("0101");break;
                case '6':stb.append("0110");break;
                case '7':stb.append("0111");break;
                case '8':stb.append("1000");break;
                case '9':stb.append("1001");break;
                case 'A':stb.append("1010");break;
                case 'B':stb.append("1011");break;
                case 'C':stb.append("1100");break;
                case 'D':stb.append("1101");break;
                case 'E':stb.append("1110");break;
                case 'F':stb.append("1111");break;
                default:break;
            }
        }
        return stb.toString();
    }
}

二、　　

import java.util.HashMap;
import java.util.Scanner;

public class Test {
    static String[] Binary = {"0000","0001","0010","0011","0100","0101","0110","0111","1000","1001","1010","1011","1100","1101","1110","1111"};
    static String[] Octal = {"000","001","010","011","100","101","110","111"};
    static HashMap<String, String> map;
    /*
     * Binary是0-15的二进制数的数组
     * Octal是0-7的二进制数的水族
     * map是0-7与其二进制数的映射
     */

    //16进制转2进制方法
    public static String HexToBinary(String hex){
        StringBuffer binary = new StringBuffer();
        for ( int i = 0 ; i < hex.length() ; i++){
            char ch = hex.charAt(i);
            if ( ch >= '0' && ch <= '9' ){
                binary.append(Binary[(int)(ch-'0')]);
            }else{
                binary.append(Binary[(int)(ch-'A')+10]);
            }
        }
        return binary.toString();
    }

    //2进制转8进制方法
    public static String BinaryToOctal(String binary){
        StringBuffer octal = new StringBuffer();
        int reminder = binary.length() % 3;
        //sub为2进制数需要补充0的个数
        int sub = 0;
        if ( reminder != 0)
            sub = 3 - reminder;
        StringBuffer tmp = new StringBuffer();
        for ( int i = 0 ; i < sub ; i++){
            tmp.append("0");
        }
        tmp.append(binary);
        for ( int i = 0 ; i <= tmp.length() - 3 ; i += 3){
            String substring = tmp.substring(i, i+3);
            octal.append(map.get(substring));
        }
        int index = FindZero(octal);
        /*
         * 如果index为-1，则8进制数前面没有0，
         * 反之则有0，取无0的子字符串。
         */
        if(index == -1)
            return octal.toString();
        else
            return octal.substring(index+1).toString();
    }

    //统计8进制数的字符串的最前面连续0的个数，返回的最后一个0的下标
    public static int FindZero(StringBuffer octal){
        int index = -1;
        for ( int i = 0 ; i < octal.length() ; i++){
            if ( octal.charAt(i) == '0'){
                index = i;
                continue;
            }else{      
                break;
            }
        }
        return index;
    }

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner in = new Scanner(System.in);
        map = new HashMap<String, String>();
        for ( int i = 0 ; i < Octal.length ; i++){
            char[] ch = new char[1];
            ch[0] = (char) (i+'0');
            String str = new String(ch);
            map.put(Octal[i], str);
        }
        int num = in.nextInt();
        for (int i = 0 ; i < num ; i++){
            String hex = in.next();
            String binary = HexToBinary(hex);
            String octal = BinaryToOctal(binary);
            System.out.println(octal);
        }
        in.close();
    }

}

三、　　

import java.util.Scanner;

public class Test1 {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		String[] st = new String[n];
		for (int i = 0; i < n; i++) {
			st[i] = sc.next();
		}
		sc.close();
		for (int i = 0; i < n; i++) {
			String str1 = ttos(st[i]);

			int len_str1 = str1.length();
			if (len_str1 % 3 == 1)
				str1 = "00" + str1;
			else if (len_str1 % 3 == 2)
				str1 = "0" + str1;
			ttoe(str1);
			System.out.println();
		}

	}

	public static String ttos(String str) {
		int len_str = str.length();
		StringBuilder str2 = new StringBuilder();
		for (int i = 0; i < len_str; ++i) {
			switch (str.charAt(i)) {
			case '0':
				str2.append("0000");
				break;
			case '1':
				str2.append("0001");
				break;
			case '2':
				str2.append("0010");
				break;
			case '3':
				str2.append("0011");
				break;
			case '4':
				str2.append("0100");
				break;
			case '5':
				str2.append("0101");
				break;
			case '6':
				str2.append("0110");
				break;
			case '7':
				str2.append("0111");
				break;
			case '8':
				str2.append("1000");
				break;
			case '9':
				str2.append("1001");
				break;
			case 'A':
				str2.append("1010");
				break;
			case 'B':
				str2.append("1011");
				break;
			case 'C':
				str2.append("1100");
				break;
			case 'D':
				str2.append("1101");
				break;
			case 'E':
				str2.append("1110");
				break;
			case 'F':
				str2.append("1111");
				break;
			default:
				break;
			}
		}
		return str2.toString();
	}

	public static void ttoe(String str2) {
		int len = str2.length();
		int a;
		a = (str2.charAt(0) - '0') * 4 + (str2.charAt(1) - '0') * 2 + (str2.charAt(2) - '0');
		if (a != 0)
			System.out.print(a);
		for (int i = 3; i <= len - 2; i += 3) {
			a = (str2.charAt(i) - '0') * 4 + (str2.charAt(i + 1) - '0') * 2 + (str2.charAt(i + 2) - '0');
			System.out.print(a);
		}
	}
}