我正在用java编写自己的大整数类,无需导入,并且需要一种方法来将字符串表示的任何大小的数字加倍。我的代码现在可以工作,但是一旦数字变得越来越大,就开始需要很长时间。我本质上创建了两个数组:主数组和倒计时数组,它们都是以相同的方式开始的。然后,我运行一个 while 循环,向上递增主数组,向下递增倒计时数组。当倒计时数组达到“0”时,我终止循环,结果是一个新数组,其中新数字的大小增加了一倍。当然,我有 if 语句检查数组是否需要更改十的位置等...这就是我所拥有的...有什么方法可以使其更加高效和快速吗?
public static String doubleDecimalString (String main) {
String countdown = main;
String finalBuild = "";
boolean runLoop = true;
//if zero is supplied, skip all the nonsense and just return 0
//else, loop through and find the true double
//was having trobule getting single digits to double correctly so i had to hard code this for now.
if (main.equals("0")) {
return main;
} else if (main.equals("5")) {
return "10";
} else if (main.equals("6")) {
return "12";
} else if (main.equals("7")) {
return "14";
} else if (main.equals("8")) {
return "16";
} else if (main.equals("9")) {
return "18";
} else {
//Array for ORIGINAL NUMBER
int[] mainPiece = new int[main.length()+2];
int arrayLength = mainPiece.length;
for ( int i = 0; i < main.length(); i++ ) {
mainPiece[i+2] = Integer.parseInt(main.substring( i, i+1));
}
mainPiece[0] = -1;
mainPiece[1] = -1;
//Array for COUNTDOWN NUMBER
int[] countdownPiece = new int[main.length()+2];
for ( int i = 0; i < main.length(); i++ ) {
countdownPiece[i+2] = Integer.parseInt(main.substring( i, i+1));
}
countdownPiece[0] = -1;
countdownPiece[1] = -1;
while ( runLoop ) {
//Increment and decrement the two arrays
mainPiece[arrayLength-1] += 1;
countdownPiece[arrayLength-1] -= 1;
//UPDATE MAIN ARRAY
if ( mainPiece[arrayLength-1] == 10 ) {
for (int x = arrayLength-1; x > 0; x--) {
if ( (mainPiece[x] == 10) && (mainPiece[x-1] != 9) ) {
mainPiece[x] = 0;
mainPiece[x -1] += 1;
} else if ( (mainPiece[x] == 10) && (mainPiece[x-1] == 9) ) {
mainPiece[x] = 0;
mainPiece[x -1] += 1;
x = arrayLength;
}
if ( (mainPiece[2] == 10) ) {
mainPiece[1] = 1;
mainPiece[2] = 0;
}
}
} // end main array
//UPDATE SIDE ARRAY
if ( countdownPiece[arrayLength-1] == -1 ) {
for (int x = arrayLength-1; x > 0; x--) {
if ( (countdownPiece[x] == -1) && (countdownPiece[x-1] > 0) && (x > 1) ) {
countdownPiece[x] = 9;
countdownPiece[x -1] -= 1;
} else if ( (countdownPiece[x] == -1) && (countdownPiece[x-1] == 0) && (x > 1) ) {
countdownPiece[x] = 9;
countdownPiece[x -1] -= 1;
x = arrayLength;
}
}
} //end side array
//tests whether the pieces need to be switched to -1 for scanning
for (int x = 0; x < arrayLength - 1; x++) {
if ( (countdownPiece[x] == -1 ) && (countdownPiece[x+1] == 0 ) ) {
countdownPiece[x+1] = -1;
}
}
//if the side array has reached "0" then the loop will stop and the main array will return the new doubled value
if ( (countdownPiece[arrayLength-1] == -1) && (countdownPiece[arrayLength-2] == -1) ) {
break;
}
} //end while loop
//transform array into string
finalBuild = "";
for (int T = 0; T < arrayLength; T++) {
finalBuild += (mainPiece[T] != -1) ? mainPiece[T] : "";
}
return finalBuild;
}
}
最佳答案
像这样的东西怎么样(它基本上是乘以二并考虑进位):
private String doubleNumber(String number)
{
int doubleDig = 0;
int carry = 0;
StringBuilder sb = new StringBuilder();
for (int i = number.length() - 1; i >= 0; --i)
{
char c = number.charAt(i);
int origNum = Character.getNumericValue(c);
doubleDig = origNum * 2 + carry;
carry = doubleDig / 10;
doubleDig = doubleDig % 10;
sb.append(doubleDig);
}
if (carry > 0)
{
sb.append(carry);
}
return sb.reverse().toString();
}
显然这只能处理整数。
关于java - 双倍十进制字符串,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29441786/