freeCodeCamp/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/100-doors.md

id, title, challengeType, forumTopicId, dashedName

id	title	challengeType	forumTopicId	dashedName
594810f028c0303b75339acb	100 门	5	302217	100-doors

--description--

连续 100 个门都是最初关闭的。 You make 100 passes by the doors. The first time through, visit every door and 'toggle' the door (if the door is closed, open it; if it is open, close it). The second time, only visit every 2nd door (i.e., door #2, #4, #6, ...) and toggle it. The third time, visit every 3rd door (i.e., door #3, #6, #9, ...), etc., until you only visit the 100th door.

--instructions--

Implement a function to determine the state of the doors after the last pass. Return the final result in an array, with only the door number included in the array if it is open.

--hints--

getFinalOpenedDoors should be a function.

assert(typeof getFinalOpenedDoors === 'function');

getFinalOpenedDoors should return an array.

assert(Array.isArray(getFinalOpenedDoors(100)));

getFinalOpenedDoors should produce the correct result.

assert.deepEqual(getFinalOpenedDoors(100), solution);

--seed--

--after-user-code--

const solution = [1, 4, 9, 16, 25, 36, 49, 64, 81, 100];

--seed-contents--

function getFinalOpenedDoors(numDoors) {

}

--solutions--

function getFinalOpenedDoors(numDoors) {
  // this is the final pattern (always squares).
  // thus, the most efficient solution simply returns an array of squares up to numDoors).
  const finalState = [];
  let i = 1;
  while (Math.pow(i, 2) <= numDoors) {
    finalState.push(Math.pow(i, 2));
    i++;
  }
  return finalState;
}