英文:
How can I turn a large integer into an array of successive two-digit integers?
问题
我想将一个n位整数j（其中n是2的幂次）转换为一个int[]数组，其中每个int[i]是由j的连续数字组成的两位整数。
所以如果
int j = 357642466853;int[] digArray = {35, 76, 42, 46, 68, 53};
我开始尝试以下方法，但陷入了困境。如果我使用String.valueOf或.charAt，它最终会将这些相应的数据类型发送到数组，尽管我希望它是一个int[]数组。
int[] digArray = new int[intLength/2]; //intLength只是nfor(int i = 0; i < intLength; i += 2) {digArray[i/2] = (String.valueOf(j).charAt(i)); //注意这还不完整，因为我甚至不知道如何在第二位数字上进行连接。}
也许我需要创建一个string[]数组，然后在完成后将其转换为int[]数组？
谢谢。
英文:
I want to take an n-digit integer j (where n is a power of two) and turn it into an int[ ] where each int[ i ] is a two-digit integer set by the successive digits of j.
So if
int j = 357642466853
int[ ] digArray = {35, 76, 42, 46, 68, 53}
I started down the following route but have gotten bogged down. If I use String.valueOf or .charAt it ends up sending those respective data types to the array, even though I want it to be an int[].
int[] digArray = new int[intLength/2] //intLength is just nfor(int i = 0; i < intLength; i += 2) {digArray[i/2] = (String.valueOf(j).charAt(i)) //Note this isn't complete as I don't even know how to concatenate on the second digit.}
Possibly I need to make a string[] and then once I'm done convert it into an int[]?
Thanks.
答案1
得分: 2
你可以在这里使用取模运算：
long j = 357642466853L;long[] digArray = new long[(int)(Math.log10(j) / 2) + 1];for (int i=digArray.length - 1; i >= 0; --i) {digArray[i] = j % 100L;j /= 100;}System.out.println(Arrays.toString(digArray));
这将打印出：
[35, 76, 42, 46, 68, 53]
英文:
You could use the modulus here:
long j = 357642466853L;long[] digArray = new long[(int)(Math.log10(j) / 2) + 1];for (int i=digArray.length - 1; i >= 0; --i) {digArray[i] = j % 100L;j /= 100;}System.out.println(Arrays.toString(digArray));
This prints:
[35, 76, 42, 46, 68, 53]
答案2
得分: 1
另一种实现方式如下：
for (long v : new long[]{ 357642466853L, 1234L, 29228912L}) {System.out.println(v + " " + Arrays.toString(longToGroups(v)));}public static long[] longToGroups(long val) {int exp = (int)Math.log10(val);exp -= (exp % 2); // 调整为奇数位长的情况long start = (long)(Math.pow(10, exp));return LongStream.iterate(start, i -> i > 0, i -> i / 100).map(i -> (val / i) % 100).toArray();}
输出结果为：
357642466853 [35, 76, 42, 46, 68, 53]1234 [12, 34]29228912 [29, 22, 89, 12]
英文:
And yet another way to do it.
for (long v : new long[]{ 357642466853L, 1234L,29228912L}) {System.out.println(v + " " + Arrays.toString(longToGroups(v)));}public static long[] longToGroups(long val) {int exp = (int)Math.log10(val);exp -= (exp % 2); // adjust for odd length longslong start =(long)(Math.pow(10,exp));return LongStream.iterate(start,i -> i > 0, i -> i / 100).map(i -> (val / i) % 100).toArray();}
Prints
357642466853 [35, 76, 42, 46, 68, 53]1234 [12, 34]29228912 [29, 22, 89, 12]</details>
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