알고리즘 문제풀이 3

hyunju-song·2020년 10월 2일
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telephoneword 문제

recursion 활용한 또 다른 문제
이것도 가위바위보 문제의 변형으로 생각하고 풀면 될듯하다.

/*
 * Each number key on a standard phone keypad has a set of Latin letters written on
 * it as well: http://en.wikipedia.org/wiki/File:Telephone-keypad2.svg
 *
 * Businesses often try to come up with clever ways to spell out their phone number
 * in advertisements to make it more memorable. But there are a lot of combinations!
 *
 * Write a function that takes up to four digits of a phone number, and
 * returns a list of all of the words that can be written on the phone with
 * that number. (You should return all permutations, not only English words.)
 *
 * Example:
 *   telephoneWords('2745');
 *   => ['APGJ',
 *        'APGK',
 *        'APGL',
 *        ..., // many many more of these
 *        'CSIL']
 *
 * Tips:
 *   - Phone numbers are strings! (A phone number can start with a zero.)
 *   - The digits 0 and 1 do not have letters associated with them, so they should be left as numbers.
 *   - Don't return every combination of those digits in any order, just the order given.
 *
 *  Extra credit: There's a list of English dictionary words at /usr/share/dict/words .
 *  Why not filter your results to only return words contained in that file?
 *

var phoneDigitsToLetters = {
  0: "0",
  1: "1",
  2: "ABC",
  3: "DEF",
  4: "GHI",
  5: "JKL",
  6: "MNO",
  7: "PQRS",
  8: "TUV",
  9: "WXYZ",
};
*/
var telephoneWords = function (digitString) {
  let results = [];

  function recur(res2, count) {
    if (count === digitString.length) {
      results.push(res2.join(""));
      //가위바위보랑 달리, 각 문자들을 문자열로 합해줘야해서 join을 써줬다.
    } else {
      let i = count; 
      //가위바위보는 for문을 써서, 재귀함수 돌때마다 다시 첫 문자들을 합쳐줬는데, 해당 문제는 주어진 인자에서 다시 앞으로 돌아가면 안된다. 
      let chars = phoneDigitsToLetters[digitString[i]].split("");
      chars.forEach((ele) => {
        recur(res2.concat(ele), count + 1);
      });
    }
  }
  recur([], 0);
  return results;
};
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