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标题: 昨天看见的题,我用最复杂的方法写的,求指正 [打印本页]

作者: a623562486 时间: 2016-9-2 21:10
标题: 昨天看见的题,我用最复杂的方法写的,求指正
/*
需求:输入阿拉伯数字,输出对应的中文大写格式
例如:11000,输出壹万壹仟
*/
/*
思路:1.首先要获取输入数字的各位数字及幂数,并对应相应汉字,考虑使用switch方法
   2.抽取进位方法,抽取拾/佰/仟/万/亿并赋值给对应幂数
*/
import java.util.Scanner;
class Chinese_Exchange {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.println("请输入一个整数,因int取值范围限制,最大取10位且小于2147483647");
int a = sc.nextInt();
int[] arr =new int[13];
String[] s = new String[13];
wei(a,arr);
//bianLi(arr);
System.out.println();
Chinese(arr,s);
//bianLi(s);
System.out.println("相应大写数字为:");
fanZhuan(s);
System.out.println();
}

public static void  wei(int a,int[] arr) {
arr[0] = a % 10;
for (int i = 1;i <arr.length-1 ;i++ ) {
a /= 10;
arr[i] = a % 10;
}
}

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

public static void bianLi(String[] s) {
for (int i = 0;i < s.length ;i++ ) {
System.out.print(s[i] + "");
}
}

public static void fanZhuan(String[] s) {
for (int i = 0;i<s.length/2 ;i++ ) {
String temp = s[i];
s[i] = s[s.length-1-i];
s[s.length-1-i] = temp;
}
bianLi(s);
}
public static void Chinese(int[] arr,String[] s) {
String[] s1 = {"零","壹","贰","叁","肆","伍","陆","柒","捌","玖"};
for (int i =0 ; i < arr.length ;i++ ) {
switch (i) {
case 0:
if (arr[i] == 0) {
s[i] = "";
break;
}
s[i] = s1[arr[i]];
break;
case 1:
if (arr[i] == 0 && (s[i-1] == "零" ||s[i-1] == "")) {
s[i] = "";
break;
}else if (arr[i] == 0 ) {
s[i] = "零";
break;
}
s[i] = s1[arr[i]] + "拾";
break;
case 2:
if (arr[i] == 0 && (s[i-1] == "零" ||s[i-1] == "")) {
s[i] = "";
break;
}else if (arr[i] == 0 ) {
s[i] = "零";
break;
}
s[i] = s1[arr[i]] + "佰";
break;
case 3:
if (arr[i] == 0 && (s[i-1] == "零" ||s[i-1] == "")) {
s[i] = "";
break;
}else if (arr[i] == 0 ) {
s[i] = "零";
break;
}
s[i] = s1[arr[i]] + "仟";
break;
case 4:
if (arr[i] == 0 && (s[i-1] == "零" ||s[i-1] == "")) {
s[i] = "";
break;
}else if (arr[i] == 0 ) {
s[i] = "零";
break;
}
s[i] = s1[arr[i]] + "萬";
break;
case 5:
if (arr[i] == 0 && (s[i-1] == "零" ||s[i-1] == "")) {
s[i] = "";
break;
}else if (arr[i] == 0 ) {
s[i] = "零";
break;
}
s[i] = s1[arr[i]] + "拾";
break;
case 6:
if (arr[i] == 0 && (s[i-1] == "零" ||s[i-1] == "")) {
s[i] = "";
break;
}else if (arr[i] == 0 ) {
s[i] = "零";
break;
}
s[i] = s1[arr[i]] + "佰";
break;
case 7:
if (arr[i] == 0 && (s[i-1] == "零" ||s[i-1] == "")) {
s[i] = "";
break;
}else if (arr[i] == 0 ) {
s[i] = "零";
break;
}
s[i] = s1[arr[i]] + "仟";
break;
case 8:
if (arr[i] == 0 && (s[i-1] == "零" ||s[i-1] == "")) {
s[i] = "";
break;
}else if (arr[i] == 0 ) {
s[i] = "零";
break;
}
s[i] = s1[arr[i]] + "亿";
break;
case 9:
if (arr[i] == 0 && (s[i-1] == "零" ||s[i-1] == "")) {
s[i] = "";
break;
}else if (arr[i] == 0 ) {
s[i] = "零";
break;
}
s[i] = s1[arr[i]] + "拾";
break;
case 10:
if (arr[i] == 0 && (s[i-1] == "零" ||s[i-1] == "")) {
s[i] = "";
break;
}else if (arr[i] == 0 ) {
s[i] = "零";
break;
}
s[i] = s1[arr[i]] + "佰";
break;
case 11:
if (arr[i] == 0 && (s[i-1] == "零" ||s[i-1] == "")) {
s[i] = "";
break;
}else if (arr[i] == 0 ) {
s[i] = "零";
break;
}
s[i] = s1[arr[i]] + "仟";
break;
case 12:
if (arr[i] == 0 && (s[i-1] == "零" ||s[i-1] == "")) {
s[i] = "";
break;
}else if (arr[i] == 0 ) {
s[i] = "零";
break;
}
s[i] = s1[arr[i]] + "萬";
break;
}
}
for (int i = arr.length-2;i >= 0 ;i-- ) {
if (arr[i] != 0) {
s[i+1] = "";
break;
}
}
}
}

作者: a623562486 时间: 2016-9-2 21:12
自己觉得太复杂了,但是只学了这么多,不知道还能怎么写

作者: NewBeeCoder 时间: 2016-9-3 22:43
思路都差不多
我想的是用一个字符串数组 String[] 存{"零","十","百","千","万","十万","百万","千万"...}
            用字符串读取输入的数字根据字符串长度读取上面数组的长度
            把输入的数字转成字符数组  用switch判断转成大写
            最后拼接字符串就行了
      只是个思路还没写代码你可以试试感觉简单点

作者: xiguapi 时间: 2016-9-4 09:39
package com.itheima.Demo;

import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.Scanner;
//需求:输入阿拉伯数字,输出对应的中文大写格式
//例如:11000,输出壹万壹仟

public class Teat12 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.println("请输入一个整数:");
int b;
while (true) {
String line = sc.nextLine();
try {
b = Integer.parseInt(line);
break;
} catch (Exception e) {
try {
BigInteger big = new BigInteger(line);
System.out.println("你输入的整数过大,请重新输入:");
} catch (Exception e1) {
try {
BigDecimal bd = new BigDecimal(line);
System.out.println("你输入的是小数,请重新输入:");
} catch (Exception e2) {
System.out.println("你输入的是非法字符,请重新输入:");
}
}
}
}
jishu(b);
}

public static void jishu(int b) {
StringBuffer str = new StringBuffer();
char[] cs = (b + "").toCharArray();
String[] js = { "零", "壹", "贰", "叁", "肆", "伍", "陆", "柒", "捌", "玖" };
String[] ws = { "元整", "拾", "佰", "仟", "万", "拾", "佰", "仟", "亿", "拾" };
for (int i = cs.length - 1; i >= 0; i--) {
int a = Integer.parseInt(cs[cs.length - 1 - i] + "");

str.append(js[a / 1] + ws[i]);
}
System.out.println(str);
}
}

有点小瑕疵.......

作者: 赢无翳 时间: 2016-9-4 09:43
我想是不是可以定义两个String类型的数组一个代表从1到9的数字一个代表万千百十判断输入数字的位数来匹配万千百十分次截取第一位上的数字来获取对应的中文

作者: xiguapi 时间: 2016-9-4 09:55

xiguapi 发表于 2016-9-4 09:39
package com.itheima.Demo;

import java.math.BigDecimal;

public static void jishu(int b) {
StringBuffer str = new StringBuffer();
char[] cs = (b + "").toCharArray();
String[] js = { "零", "壹", "贰", "叁", "肆", "伍", "陆", "柒", "捌", "玖" };
String[] ws = { "元整", "拾", "佰", "仟", "万", "拾", "佰", "仟", "亿", "拾" };
for (int i = cs.length - 1; i >= 0; i--) {
int a = Integer.parseInt(cs[cs.length - 1 - i] + "");
if (a / 1 == 0 && (i == 0 || i == 4 || i == 8)) {
str.append(ws);
} else {
if (a / 1 == 0) {

} else {
str.append(js[a / 1] + ws);
}
}
}
System.out.println(str);
}

这样就好了!!!!
作者: a623562486 时间: 2016-9-5 20:50

赢无翳发表于 2016-9-4 09:43
我想是不是可以定义两个String类型的数组一个代表从1到9的数字一个代表万千百十判断输入数字的位数来匹 ...

这个是可行的,但是里面要加判断语句,因为如果涉及到很多位都是零的时候,就很麻烦
作者: a623562486 时间: 2016-9-5 20:56

xiguapi 发表于 2016-9-4 09:39
package com.itheima.Demo;

import java.math.BigDecimal;

还没学到integer类呢,诶,等几天再回来看

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