Postingan lainnya
udah 2 hai tapi gak pecah pecah,, nih masalah udah kusut pala ane gan plis bantu
var input = 'abc';
var bitmask = 0;
var characters = input.split('');
var running = [];
function permutations() {
var i;
if(running.length == characters.length) {
console.log(running.join(''));
} else {
for(i=0; i<characters.length; i++) {
if ( ((bitmask>>i)&1) == 0 ) {
running.push(characters[i]);
bitmask |= (1<<i);
permutations();
running.pop();
}
}
}
}
permutations();
gimana agar keluarnya kayak gini gan,, abc acb bac bca cab cba
yang salah dari yang ane buat di atas mastah tolong bantu ane udah 2 hari ampe ane gebukin pala sendiri tapi gx kepecah di mana sih salahnya mastah,,??
sekian terima kasih
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2 Jawaban:
function permut(string) {
if (string.length < 2) return string; // This is our break condition
var permutations = []; // This array will hold our permutations
for (var i=0; i<string.length; i++) {
var char = string[i];
// Cause we don't want any duplicates:
if (string.indexOf(char) != i) // if char was used already
continue; // skip it this time
//Note: you can concat Strings via '+' in JS
var remainingString = string.slice(0,i) + string.slice(i+1,string.length);
for (var subPermutation of permut(remainingString))
permutations.push(char + subPermutation)
}
return permutations;
}
console.log(permut('abc'));
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wehhh.. mkasih bang cara abang juga lebih simple makasih dah bg,, kalo ane cewek udh ane kasih kode dah untung nya ane cowok :v
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