在我一直在努力的项目中，我遇到了一个问题，我需要在Integer附加一个数字。问题是您无法通过Integer或Integer.append(someArrayOfNumbers)这样的简单解决方案附加到someDigit.append(aNumber).，并且其中并没有任何好有关它的信息。所以我想知道一些算法和＆＃34; 这样的＆＃34;。
答案 0 :(得分：2)
一种解决方法是将int乘以10并添加数字，或者如果int为负数，则减去数字：
pow(10, log10(yourDigits) + 1)如果要追加任意数量的数字，请将整数乘以Integer.parseInt(Integer.toString(yourInt) + Integer.toString(yourDigit))
。
另一个解决方案是将int转换为字符串并在字符串末尾添加数字，并将其转换回int：
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答案 1 :(得分：1)
这是我的 基础 解决方案（好吧部分我的，我从这个答案得到了一些信息，但我不得不修补， here），不使用String。我说 base 因为如果其中一个数组成员超过一位数，这个算法就无法正常工作。
public int append(int[] arrayOfIntegers) {
int total;
int length = arrayOfIntegers.length; // for quick reference
total = (int) (arrayOfIntegers[0] * Math.pow(10, length - 1)); // first item times 10 to the power of length minus 1
for (int i = 1; i < length; i++) {
total += arrayOfIntegers[i] * Math.pow(10, length - i - 1);
}
return total;
}
现在附加一个整数：
int[] array = { 1, 2, 3, 4, 5, 6, 7 };
System.out.println("The array appended: " + append(array));
输出：
The array appended: 1234567
答案 2 :(得分：0)
如果您将仅一位附加到右侧或LSB，则可以通过乘以10并添加新数字来附加...
public int append(int originalValue, int numberToAppend ) {
return originalValue*10+ numberToAppend;
}
您可以使用首先使用字符串连接来执行更多数字
public int append(Integer originalValue, Integer numberToAppend ) {
String temp = originalValue.toString()+ numberToAppend.toString();
return Integer.parseInt(temp);
}