我的家庭作业让我检查用户输入的数字中所有可能的循环符号。我已将输入发送到数组中，但我不确定如何启动循环。我如何编辑此循环以不多次显示相同的数字？第一次发帖，格式不对，请见谅。
// example of user input
var permutation = [ 9,2,3,7,4,1,8,6,5 ] ;
// add zero to match index with numbers
permutation.unshift(0) ;
// loop to check for all possible permutations
for (var i = 1; i < permutation.length -1; i++)
{
var cycle = [];
var currentIndex = i ;
if ( permutation [ currentIndex ] == i )
cycle.push(permutation [currentIndex]);
while ( permutation [ currentIndex ] !== i )
{
cycle.push( permutation [currentIndex]);
currentIndex = permutation [ currentIndex ] ;
}
// display in console
console.log(cycle);
}
我回答过类似的问题here https://stackoverflow.com/questions/40623774/array-arrangement-permutation/40625585#40625585。这个想法是使用递归函数。
这是一个示例：
var array = ['a', 'b', 'c'];
var counter = 0; //This is to count the number of arrangement possibilities
permutation();
console.log('Possible arrangements: ' + counter); //Prints the number of possibilities
function permutation(startWith){
startWith = startWith || '';
for (let i = 0; i < array.length; i++){
//If the current character is not used in 'startWith'
if (startWith.search(array[i]) == -1){
//If this is one of the arrangement posibilities
if ((startWith + array[i]).length == array.length){
counter++;
console.log(startWith + array[i]); //Prints the string in console
}
//If the console gives you "Maximum call stack size exceeded" error
//use 'asyncPermutation' instead
//but it might not give you the desire output
//asyncPermutation(startWith + array[i]);
permutation(startWith + array[i]); //Runs the same function again
}
else {
continue; //Skip every line of codes below and continue with the next iteration
}
}
function asyncPermutation(input){
setTimeout(function(){permutation(input);},0);
}
}
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