Question

我对编码还很陌生，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。我一次只能++或-一个数字。

这是挑战：

不允许对数字重新排序例如，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
}

Answer 1

我认为您应该创建一个有效地使用一位数字的函数，而该数字要比另一个数字大（又需要递减），则可以这样做：

const incrementWhileNeeded = (target, currentValue) =>
  Math.abs(target - currentValue)

然后，您需要将数字分解为数字（可以像看起来一样使用%进行数学处理，但为简单起见，将使用诸如String(num1).split('').map(Number) 451，并将其更改为[4, 5, 1]。

然后，下一步是将该函数（incrementWhileNeeded）映射到每个数字：只关注第一个数字（然后应用forEach或.map将该函数应用于所有数字。

所以看起来像这样： firstNumberArray.map(incrementWhileNeeded)

如您所解释的，[1, 0, 2]会回复您。

然后.reduce（）这样，您可以获取计数的总和。因此，这会将使用[1,0,2].reduce((accumulator, current) => accumulator + current)减少到3。

因此要获得全部功能：

const incrementWhileNeeded = (target, currentValue) =>
      Math.abs(target - currentValue)

const calculateMinimumMoves = (fullNumber, targetNumber) => {
      const numArray = String(fullNumber).split('').map(Number)
      const targetArray = String(targetNumber).split('').map(Number)
      const diffArray = numArray.map((currentElement, targetArray[index]) => incrementWhileNeeded(currentElement, targetArray[index])
      return diffArray.reduce((accumulator, current) => accumulator + current, 0)
}

const minimumMoves = (array1, array2) =>
      array1.reduce((accumulator, current, index) =>
            accumulator + calculateMinimumMoves(current, array2[index]),
            0)

Answer 2

要仅将changina一串数字的计数获取到另一个数字，可以在某个位置添加数字的绝对增量。

function count(a, b) {
    return Array.from(a).reduce((s, v, i) => s + Math.abs(v - b[i]), 0);
}

console.log(count('123', '321'));

Answer 3

签出此代码：

a = [1234, 4321]
b = [2345, 3214]

function minimumMoves(a, 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
    }
}

如果您需要对此代码进行快速验证，请导航Here。

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, i.e, either numMoves1 or numMoves2
        }
        System.out.println(numMoves1 + " & " + numMoves2);
    }
}

我希望此算法有帮助;）

Answer 4

//This code works....
// Check this out ....
public class Main
{
    public static void main(String[] args) {
        Integer[] a = {1234, 4321};
        Integer[] m = {2345, 3214};
        Integer numMoves1 = 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];
           while(num1>0)
            {
                digit1 = num1 % 10;
                digit2 = num2 % 10;
                numMoves1 += Math.abs(digit1-digit2);
                num1 = (num1 - digit1) / 10;
                num2 = (num2 - digit2) / 10;
            }
        }
        System.out.println(numMoves1);
    }
}

Answer 5

这是找到最小移动以匹配两个不同数组的每个元素的解决方案。

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]; 
        num2 = m[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;
}

console.log(minimumMoves(a, m));

最小位数更改以使两个数组具有相同的值

5 个答案: