我设定的任务是创建一个给定数组数的所有排列/ 4位数的数组：[1,2,3,4,5,6,7,8,9]。不能重复数字，因为每个值必须是唯一的。下面是我的解决方案，但我正在努力将递归应用于该过程。我希望它具有适应性，因此如果条件发生变化，即函数必须产生5位数字甚至6位数的所有组合，因为需要更改少量代码并且添加递归将容易实现。如下所示，代码可以正常工作，但如果条件发生变化，则需要更多嵌套for循环。 我正在寻找递归解决方案。这似乎不是一个好的解决方案，但任何建议都会非常感激。我在网上看到很多关于创建4P4或5P5但不是9P5样式的解决方案。我试图应用堆的算法，但没有成功。
function arrayCreate((availableNumbers, userNumberArray)) {
var possibleValues = []; //empty array to house all the possible combination of values that the user could enter i.e. 1234 to 9876
var numberOfPermutations = (factorial(availableNumbers.length) / factorial(availableNumbers.length - userNumberArray.length));
var adding = true;
var firstDigit, secondDigit, thirdDigit, forthDigit =0;
var possibleDigitValue = "";
while (adding === true) {
for (var i = 0; i < availableNumbers.length; i++) {
firstDigit = availableNumbers[i];
availableNumbers.splice(i, 1);
for (var j = 0; j < availableNumbers.length; j++) {
secondDigit = availableNumbers[j];
availableNumbers.splice(j, 1);
for (var k = 0; k < availableNumbers.length; k++) {
thirdDigit = availableNumbers[k]
availableNumbers.splice(k, 1);
for (var l = 0; l < availableNumbers.length; l++) {
forthDigit = availableNumbers[l];
possibleDigitValue = (firstDigit + secondDigit + thirdDigit + forthDigit);
possibleValues.push(possibleDigitValue);
}
availableNumbers.splice(k, 0, thirdDigit);
}
availableNumbers.splice(j, 0, secondDigit);
}
availableNumbers.splice(i, 0, firstDigit);
if (possibleValues.length >= numberOfPermutations) {
adding = false;
}
}
console.log(possibleValues);
return possibleValues;
}
}
arrayCreate([1,2,3,4,5,6,7,8,9],[0,0,0,0]);
var userNumberArray = ['0', '0', '0', '0']; //empty array of 0's as this value is not allowed, this array will store the computers auto-generated number
var availableNumbers = ['1', '2', '3', '4', '5', '6', '7', '8', '9'] //array of available numbers to be picked and added to the computerNumber array
//this function is used later to calculate the possible permutations of combinations of user guess
function factorial(x) {
if (x === 0) { return 1; }
else{
return x * factorial(x-1);
}
}
function arrayCreate(availableNumbers, userNumberArray) {
var possibleValues = []; //empty array to house all the possible combination of values that the user could enter i.e. 1234 to 9876
var numberOfPermutations = (factorial(availableNumbers.length) / factorial(availableNumbers.length - userNumberArray.length));
var adding = true;
var firstDigit, secondDigit, thirdDigit, forthDigit =0;
var possibleDigitValue = "";
while (adding === true) {
for (var i = 0; i < availableNumbers.length; i++) {
firstDigit = availableNumbers[i];
availableNumbers.splice(i, 1);
for (var j = 0; j < availableNumbers.length; j++) {
secondDigit = availableNumbers[j];
availableNumbers.splice(j, 1);
for (var k = 0; k < availableNumbers.length; k++) {
thirdDigit = availableNumbers[k]
availableNumbers.splice(k, 1);
for (var l = 0; l < availableNumbers.length; l++) {
forthDigit = availableNumbers[l];
possibleDigitValue = (firstDigit + secondDigit + thirdDigit + forthDigit);
possibleValues.push(possibleDigitValue);
}
availableNumbers.splice(k, 0, thirdDigit);
}
availableNumbers.splice(j, 0, secondDigit);
}
availableNumbers.splice(i, 0, firstDigit);
if (possibleValues.length >= numberOfPermutations) {
adding = false;
}
}
return possibleValues;
}
}
console.log(arrayCreate(availableNumbers, userNumberArray));
答案 0 :(得分：2)
您可以通过迭代项目并检查之前是否已选择项目来采取递归方法。如果没有，则取出该项目并检查零件阵列的长度。
如果它具有所需的长度，则将部分数组放入结果中。
如果没有，则迭代给定的数组并移交部件数组。
function arrayCreate(array, size) {
var result = [];
array.forEach(function iter(parts) {
return function (v) {
var temp = parts.concat(v);
if (parts.includes(v)) {
return;
}
if (temp.length === size) {
result.push(temp);
return;
}
array.forEach(iter(temp));
}
}([]));
return result;
}
console.log(arrayCreate([1, 2, 3, 4, 5, 6, 7, 8, 9], 4).map(a => a.join('')));
console.log(arrayCreate([1, 2, 3, 4, 5, 6, 7, 8, 9], 5).map(a => a.join('')));
console.log(arrayCreate([1, 2, 3, 4, 5, 6, 7, 8, 9], 6).map(a => a.join('')));.as-console-wrapper { max-height: 100% !important; top: 0; }
使用分隔函数iter。
function arrayCreate(array, size) {
function iter(parts) {
return function (v) {
var temp = parts.concat(v);
if (parts.includes(v)) {
return;
}
if (temp.length === size) {
result.push(temp);
return;
}
array.forEach(iter(temp));
}
}
var result = [];
array.forEach(iter([]));
return result;
}
console.log(arrayCreate([1, 2, 3, 4, 5, 6, 7, 8, 9], 4).map(a => a.join('')));
console.log(arrayCreate([1, 2, 3, 4, 5, 6, 7, 8, 9], 5).map(a => a.join('')));
console.log(arrayCreate([1, 2, 3, 4, 5, 6, 7, 8, 9], 6).map(a => a.join('')));.as-console-wrapper { max-height: 100% !important; top: 0; }
答案 1 :(得分：-1)
任何规模和任何选项的解决方案
const makeUniqPermutations = (size, options) => {
if (size > options.length) {
throw new Error('Cannot make uniq permutations with that size and options');
}
if (size === 0) {
return [''];
}
const permutations = options.reduce((acc, option, index) => {
const restSize = size - 1;
const restOptions = [
...options.slice(0, index),
...options.slice(index + 1),
];
const restPermutations = makeUniqPermutations(restSize, restOptions);
const newPermutations = restPermutations.map(permutation => `${option}${permutation}`);
return [...acc, ...newPermutations];
}, [])
return permutations;
}
const options = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
const size = 4;
console.log(makeUniqPermutations(size, options));