我对编码很陌生,我正在尽力而为,但是经过数小时的研究,我仍然无法弄清这一点.我正在尝试通过最少的移动使这两个单独的数组相同.我只能或-一次只能有一个号码.
这是挑战：
不允许对数字进行重新排序例如,考虑两个数组：Andrea的[123,543]和Maria的[321,279].对于第一个数字,Andrea可以将1递增两次以达到3.2已经相等.最后,她将3减2等于1.花了4步才能达到目标.对于第二个整数,她递减5 3次,递增4 3次和3 6次.花了12步来转换第二个数组元素.总共花了16步来转换组成整个数组的两个值.
let a = [1234,4321]
let m = [2345,3214]
function minimumMoves(a,m) {
// Write your code here
let numMoves = 0;
let num1 = '' ;
let num2 = '' ;
let digit1 = '';
let digit2= '';
for (let i = 0; i < a.length; i++)
{
num1 = a[i];
while (num1 != 0) {
digit1 = num1 % 10;
digit2 = num2 % 10;
num1 = Math.trunc(num1 / 10);
num2 = Math.trunc(num2 / 10);
numMoves = numMoves + Math.abs(digit1 - digit2);
}
}
return numMoves
}
最佳答案
检查此代码：
原文链接：https://www.f2er.com/js/531223.htmla = [1234,4321]
b = [2345,m) {
let numMoves1 = 0,numMoves2 = 0;
let num1 = '',num2 = '';
let digit1 = '',digit2 = '';
//Forward
for (let i = 0 ; i < a.length ; i++)
{
num1 = a[i];
num2 = m[i];
for (let j = 0 ; j < a.length ; j++)
{
digit1 = num1 % 10;
digit2 = num2 % 10;
numMoves1 += Math.abs(digit1-digit2);
num1 = (num1 - digit1) / 10;
num2 = (num2 - digit2) / 10;
}
}
//Backward
for (let i = 0 ; i < a.length ; i++)
{
num1 = m[i];
num2 = a[i];
for (let j = 0 ; j < a.length ; j++)
{
digit1 = num1 % 10;
digit2 = num2 % 10;
numMoves2 += Math.abs(digit1-digit2);
num1 = (num1 - digit1) / 10;
num2 = (num2 - digit2) / 10;
}
}
if (numMoves1>numMoves2)
{
//Answer is numMoves1
} else if (numMoves1<numMoves2)
{
//Answer is numMoves2
} else {
//Answer is any one,i.e,either numMoves1 or numMoves2
}
}
然后粘贴以下代码：
/******************************************************************************
Online Java Compiler.
Code,Compile,Run and Debug java program online.
Write your code in this editor and press "Run" button to execute it.
*******************************************************************************/
public class Main
{
public static void main(String[] args) {
Integer[] a = {1234,4321};
Integer[] m = {2345,3214};
Integer numMoves1 = 0,numMoves2 = 0;
Integer num1 = 0,num2 = 0;
Integer digit1 = 0,digit2 = 0;
//Forward
for (Integer i = 0 ; i < a.length ; i++)
{
num1 = a[i];
num2 = m[i];
for (Integer j = 0 ; j < a.length ; j++)
{
digit1 = num1 % 10;
digit2 = num2 % 10;
numMoves1 += Math.abs(digit1-digit2);
num1 = (num1 - digit1) / 10;
num2 = (num2 - digit2) / 10;
}
}
//Backward
for (Integer i = 0 ; i < a.length ; i++)
{
num1 = m[i];
num2 = a[i];
for (Integer j = 0 ; j < a.length ; j++)
{
digit1 = num1 % 10;
digit2 = num2 % 10;
numMoves2 += Math.abs(digit1-digit2);
num1 = (num1 - digit1) / 10;
num2 = (num2 - digit2) / 10;
}
}
if (numMoves1>numMoves2)
{
//Answer is numMoves1
} else if (numMoves1<numMoves2)
{
//Answer is numMoves2
} else
{
//Answer is any one,either numMoves1 or numMoves2
}
System.out.println(numMoves1 + " & " + numMoves2);
}
}
我希望这个算法有帮助
