可以每一位用一个Int表示,存在一个int[]里面。 这个数组最大长度是num1.len + num2.len,比如99 * 99,最大不会超过10000,所以4位就够了。 这种个位在后面的,不好做(10的0次方,可惜对应位的数组index不是0而是n-1), 所以干脆先把string reverse了代码就清晰好多。 最后结果前面的0要清掉。 代码如下: ...
+String[] args +static void main(String[]) } MULTIPLY { +int multiply(int a, int b) } MAIN ||--o| MULTIPLY : "利用" 结尾 通过以上的步骤,你已经成功实现了一个简单的Java乘法运算。学习编程是一项循序渐进的工作,希望你能够在实践中不断积累经验。开始动手实现吧,乘法不仅是数学中的基础运算,更...
把结果放在一个数组里,用offset来移动起始位置。 1publicclassSolution {2publicString multiply(String num1, String num2) {3if(num1 ==null|| num1.length() ==0|| num2 ==null|| num2.length() ==0)return"";4int[] result =newint[num1.length() + num2.length()];//maximum length56int...
privateStringadd(Strings1,Strings2) { Stringresult=""; intcarry=0; inti=s1.length()-1, j=s2.length()-1; for(; i>=0||j>=0; i--, j--) { intl=(i>=0)?Integer.parseInt(String.valueOf(s1.charAt(i))):0; intr=(j>=0)?Integer.parseInt(String.valueOf(s2.charAt(j))):0; ...
(String[]args){Scannerin=newScanner(System.in);// Creating a Scanner object to read input from the user// Asking the user to input the first numberSystem.out.print("Input the first number: ");intn1=in.nextInt();// Reading the first integer input from the user// Asking the user to...
StringBuffer StringBuilder StringIndexOutOfBoundsException SuppressWarnings SuppressWarningsAttribute Thread Thread.IUncaughtExceptionHandler Thread.State ThreadDeath ThreadGroup ThreadLocal Throwable TypeNotPresentException UnknownError UnsatisfiedLinkError UnsupportedClassVersionError ...
Java实现 1classSolution {2publicString multiply(String num1, String num2) {3intm =num1.length();4intn =num2.length();5int[] res =newint[m +n];6for(inti = m - 1; i >= 0; i--) {7for(intj = n - 1; j >= 0; j--) {8intproduct = (num1.charAt(i) - '0') * ...
add string data to IList collection Add strings to list and expiry each item in certain period of time add text file data into arraylist Add Text to a Textbox without removing previous text Add Two Large Numbers Using Strings - Without Use of BigInt Add user properties settings at run time...
A=(A 1 ,A 2 ,…,A N ), and an integer C. Find the maximum possible sum of the elements in A after performing the following operation at most once: Specify integers l and r such that 1≤l≤r≤N, and multiply each of A l ,A l+1 ,…,A r by C. ...
multiply(new BigDecimal(String.valueOf(factor))) Java documentation forjavax.xml.datatype.Duration.multiply(int). Portions of this page are modifications based on work created and shared by theAndroid Open Source Projectand used according to terms described in th...