每日一道算法题 - KaprekarsConstant(hard-1)

虽然都是很简单的算法，每个都只需5分钟左右，但写起来总会遇到不同的小问题，希望大家能跟我一起每天进步一点点。
更多的小算法练习，可以查看我的文章。

规则

Using the JavaScript language, have the function KaprekarsConstant(num) take the num parameter being passed which will be a 4-digit number with at least two distinct digits. Your program should perform the following routine on the number: Arrange the digits in descending order and in ascending order (adding zeroes to fit it to a 4-digit number), and subtract the smaller number from the bigger number. Then repeat the previous step. Performing this routine will always cause you to reach a fixed number: 6174. Then performing the routine on 6174 will always give you 6174 (7641 - 1467 = 6174). Your program should return the number of times this routine must be performed until 6174 is reached. For example: if num is 3524 your program should return 3 because of the following steps: (1) 5432 - 2345 = 3087, (2) 8730 - 0378 = 8352, (3) 8532 - 2358 = 6174.

使用JavaScript语言，让函数KaprekarsConstant（num）获取传递的num参数，该参数将是一个4位数字，至少有两个不同的数字。
按降序和升序排列数字（添加零以使其适合4位数字），用较大的数字减去较小的数字。
然后重复上一步，直到相减结果等于固定数字：6174，返回该程序必须执行的次数。
例如：如果传入的参数为3524，你的程序应该返回3，即该程序必须执行3次才能得到6174的结果。
因为以下步骤：（1）5432 - 2345 = 3087，（2）8730 - 0378 = 8352，（3）8532 - 2358 = 6174.

function KaprekarsConstant(num) { // code goes here  return num;
}// keep this function call here
KaprekarsConstant(3524);

测试用例

Input:2111
Output:5Input:9831
Output:7

my code

function KaprekarsConstant(num) {var count = 0while (true) {var maxNum = num.toString().split('').sort((item1, item2) => item2 - item1).join('')maxNum = Number(maxNum)var minNum = num.toString().split('').sort().join('')minNum = Number(minNum)num = '0000' + (maxNum - minNum)num = num.substr(-4)count++if (num == 6174 || num == 0) break}return count
}

other code

code-1

function KaprekarsConstant(num) {const KAP = 6174;var count = 0;while (true) {var num = evaluator(num)if (num === true) {return count;}}function evaluator(num) {count++console.log('count', count);var minNumArr = num.toString().split('').sort();var maxNumArr = minNumArr.slice(0).reverse();var littleNum = parseInt(minNumArr.join(''), 10);var bigNum = parseInt(maxNumArr.join(''), 10);while (bigNum < 1000) {bigNum = bigNum * 10;}return bigNum - littleNum === KAP ? true : bigNum - littleNum;}
}

code-2

function KaprekarsConstant(num) {var count = 0;while (num != 6174) {count += 1;var numArr = num.toString().split('');while (d.length < 4) {numArr.push('0');}var smaller = numArr.sort().join('');var bigger = numArr.reverse().join('');num = bigger - smaller;}return count;
}

code-3

function KaprekarsConstant(num) { let count = 0;while (num != 6174) {let numArray = num.toString().split('').sort();let ascending = parseInt(numArray.join(''));let descending = parseInt(numArray.reverse().join(''));while (descending.toString().length < 4) {descending *= 10;}num = Math.abs(ascending - descending);count++;if (count > 999) break;  // failover}return count;
}

思路

个人思路：

本想使用递归去解决，代码量会更少更优雅，然后发现需要计算执行的次数，所以才使用while去遍历执行（其实使用闭包也可以解决，无奈题目调用就是KaprekarsConstant(xxxx)）
先把数字转成字符串再转数组，使用数组升降排序拿到最大最小值，相减之后得到结果
给相减结果补0，保证相减结果一直都是4位数
当相减结果等于6174或0时，中断while循环，返回执行次数

优化点：
1.minArr = num.toString().split('').sort(); minNum = minArr.join("")，那么maxNum可以直接使用minArr.reverse().join('')得到
2.可以把if判断内容放到while上
3.补0可以使用 num * 10 的方式，而不是字符串补零