想要打印一个单词的英文表示（介于0和999,999之间），例如1,234，我想打印为“一千二百三十四”。找到以下解决方案，但不确定第23行到第25行的逻辑含义。任何见解都值得赞赏。谢谢。
1 public static String numtostring(int num) {
2 StringBuilder sb = new StringBuilder();
3
4 // Count number of digits in num.
5 int len = 1;
6 while (Math.pow((double)10, (double)len ) < num) {
7 len++;
8 }
9
10 String[] wordarr1 = {“”,”One ”, “Two ”, “Three ”, “Four ”,
11 “Five ”, “Six ”, “Seven ”, “Eight ”,”Nine ”};
12 String[] wordarr11 = {“”, “Eleven ”, “Twelve ”, “Thirteen ”,
13 “Fourteen ”, “Fifteen ”, “Sixteen ”,
14 “Seventeen ”, “Eighteen ”, “Nineteen ”};
15 String[] wordarr10 = {“”,”Ten ”, “Twenty ”, “Thirty ”, “Forty ”,
16 “Fifty ”, “Sixty ”, “Seventy ”, “Eighty ”,
17 “Ninety “};
18 String[] wordarr100 = {“”, “Hundred ”, “Thousand ”};
19 int tmp;
20 if (num == 0) {
21 sb.append(“Zero”);
22 } else {
23 if (len > 3 && len % 2 == 0) {
24 len++;
25 }
26 do {
27 // Number greater than 999
28 if (len > 3) {
29 tmp = (num / (int)Math.pow((double)10,(double)len-2));
30 // If tmp is 2 digit number and not a multiple of 10
31 if (tmp / 10 == 1 && tmp%10 != 0) {
32 sb.append(wordarr11[tmp % 10]) ;
33 } else {
34 sb.append(wordarr10[tmp / 10]);
35 sb.append(wordarr1[tmp % 10]);
36 }
37 if (tmp > 0) {
38 sb.append(wordarr100[len / 2]);
39 }
40 num = num % (int)(Math.pow((double)10,(double)len-2));
41 len = len-2;
42 } else { // Number is less than 1000
43 tmp = num / 100;
44 if (tmp != 0) {
45 sb.append(wordarr1[tmp]);
46 sb.append(wordarr100[len / 2]);
47 }
48 tmp = num % 100 ;
49 if(tmp / 10 == 1 && tmp % 10 != 0) {
50 sb.append(wordarr11[tmp % 10]) ;
51 } else {
52 sb.append(wordarr10[tmp / 10]);
53 sb.append(wordarr1[tmp % 10]);
54 }
55 len = 0;
56 }
57 } while(len > 0);
58 }
59 return sb.toString();
60 }
答案 0 :(得分：1)
您正在谈论此代码段：
if (len > 3 && len % 2 == 0) {
len++;
}
这意味着如果len大于3（显然！）并且len是even数字。
%符号用于modulo operation。
维基百科关于模运算的定义是：
在计算中，模运算在除法后找到余数 一个数字与另一个数字（有时称为模数）。
<强>更新
关于此代码段背后的逻辑。我只会试着猜测原作者的意图。 len变量将保留数字的长度。
如果长度小于100，则作者将在} else { // Number is less than 1000块中创建所需的字符串。如果您看到更好的代码段，则永远不会使用len变量，除非他使其等于零以退出循环。
因此，对于数字＆lt; 999从未使用len。
现在，对于数字＆gt; 1000它被使用，我想作者需要更改它，以便将其用于阵列访问。这就是使用第23至25行的原因。因此，对于数字1000到9999 len是4，对于10000到99999 len是5而对于100000到999999 len是6.我想如果不改变它，则作者无法访问所需的数组值。这是我猜测为什么使用这段代码的原因。
但是，我猜测从给定的字符串开始，该方法对0到999999之间的数字应该可以正常工作。试试这个想法：
int err = 0;
for(int i = 0;i<999999;i++) {
try {
numtostring(i);
} catch (Exception e){
err++;
}
}
System.out.println(err+" ERORRS");
这将打印900000 ERRORS。对于1000的情况和从100000到999999的所有数字的情况。这两种情况在此代码中处理不正确。我发现这整个len操作非常错误，很难遵循。我不确定作者是否有其他想法。
这是一个修订版本，它似乎适用于从0到999999的数字，我发现它更容易理解。
public static String numtostring(int num) {
StringBuilder sb = new StringBuilder();
// Count number of digits in num.
int len = String.valueOf(num).length();
String[] wordarr1 = {"", "One ", "Two ", "Three ", "Four ",
"Five ", "Six ", "Seven ", "Eight ", "Nine "};
String[] wordarr11 = {"", "Eleven ", "Twelve ", "Thirteen ",
"Fourteen ", "Fifteen ", "Sixteen ",
"Seventeen ", "Eighteen ", "Nineteen "};
String[] wordarr10 = {"", "Ten ", "Twenty ", "Thirty ", "Forty ",
"Fifty ", "Sixty ", "Seventy ", "Eighty ",
"Ninety "};
int tmp;
if (num == 0) {
sb.append("Zero");
} else if (num >= 1000000) {
System.err.println("Numbers > 999999 are not supported!");
System.exit(1);
} else {
do {
// Number greater than 999
if (len > 3) {
int n = num / 1000;
sb.append(numtostring(n)).append("Thousand ");
num = num % 1000;
len -= String.valueOf(n).length();
} else { // Number is less than 1000
tmp = num / 100;
if (tmp != 0) {
sb.append(wordarr1[tmp]);
sb.append("Hundred ");
}
tmp = num % 100;
if (tmp / 10 == 1 && tmp % 10 != 0) {
sb.append(wordarr11[tmp % 10]);
} else {
sb.append(wordarr10[tmp / 10]);
sb.append(wordarr1[tmp % 10]);
}
len = 0;
}
} while (len > 0);
}
return sb.toString();
}
您还可以通过再次调用do-while numtostring()作为参数并将其附加到n来消除sb循环。
无论如何，在使用它之前先测试一下，以确保我没有忘记任何东西:)我希望我帮助你一点！
更新2
好的，所以我们讨论过的原始方法适用于长度<1的数字。 3.它适用于4位和5位数字，但不适用于6位数之一。
让我们看看如何使用len：
if (len > 3 && len % 2 == 0) {
// if len is 4 it becomes 5, if it is 5 it stays as is
// if len is 6 it becomes 7 and an exception occurs
len++;
}
...
if (len > 3) {
// puts the thousand part in tmp
// so if num is 9000, len is 5 and tmp is 9 (9000/10^3)
// if num is 99000, len is 5 and tmp is 99 (99000/10^3)
// and if num is 999000, len is 7 and tmp is 9 instead of 999 (999000/10^5)
tmp = (num / (int)Math.pow((double)10,(double)len-2));
// If tmp is 2 digit number and not a multiple of 10
// So, if tmp is 11 to 19 (num was 11000 to 19999) it enters the if
if (tmp / 10 == 1 && tmp%10 != 0) {
// if tmp is 11 tmp % 10 is 1 and the wordarr11[1] is eleven etc.
sb.append(wordarr11[tmp % 10]) ;
} else {
// if tmp is not 11 to 19 it enters here
// this means if tmp is 1 to 9 for num 1000 to 9999
// if tmp is 10 for num 10000 to 10999
// if tmp is 20 to 99 for num 20000 to 99999
// if tmp is 100 to 999 for num 100000 to 999999
// wordarr10 contains the dozens
// if tmp is 10 this will be ten, if it is 20 this will be twenty etc.
// if tmp is 1 to 9 tmp / 10 will return 0 and sb will append an empty string (wordarr10[0])
sb.append(wordarr10[tmp / 10]);
// wordarr1 contains the units
// if tmp is 1 to 9 then tmp % 10 will return the tmp as it is
// if tmp is 10 or 20 tmp % 10 will return 0 and append the empty string
// if tmp is 23 tmp % 10 will return 3 and append the word three
sb.append(wordarr1[tmp % 10]);
}
// if tmp is a positive numbers... we know it is but ok...
// we append the word hundrend if len is 2 or three, which is impossible because we are in the if(len > 3) branch
// if original len was 4 it have become 5 earlier so
// if len is 5 the len / 2 is 2 and we append the word thousand
// if len is 7 len / 2 is 3 and an out of bounds exception gets thrown
if (tmp > 0) {
sb.append(wordarr100[len / 2]);
}
// finally we remove the part of num that we have printed in order to print the rest
// so if num is 1123 then it will become 123
// or if it is 12123 it will become again 123 (because len is 5)
// if len is 7 this will fail and for example 123123 will become 23123
num = num % (int)(Math.pow((double)10,(double)len-2));
// if len is 5 then we make it three in order to run the else branch and print the rest part
// if len was 7 this would make it 5 and the same branch would run again which I guess is also wrong
len = len-2;
从这两行中或多或少地创造了整个混乱：
tmp = (num / (int)Math.pow((double)10,(double)len-2));
num = num % (int)(Math.pow((double)10,(double)len-2));
他们最好使用1000代替(int)Math.pow((double)10,(double)len-2)。然后其余部分或多或少会像其他部分一样。请参阅我的第一次更新，我在修改后的代码中这样做。
最后，还有另一个问题。正如我之前所说的那样，1000再次出现例外情况。这是因为长度计数错误。
int len = 1;
while (Math.pow((double)10, (double)len ) < num) {
len++;
}
对于1000，它将返回len = 3，但对于1001，它将返回len = 4。对于10000，它将返回len = 4，但对于10001，它将返回len = 5。
如果您有特定问题，请询问：）