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김승우 3주차 과제 구현 #2
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hannut91
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Sep 26, 2023
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for (let i = 0; i < array.length; i++) { | ||
for (let j = 0; j < array.length - 1 - i; j++) { | ||
if (array[j] > array[j + 1]) { | ||
exchangeCount += 1; | ||
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[array[j + 1], array[j]] = [array[j], array[j + 1]]; | ||
} | ||
} | ||
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if (i === 0 && exchangeCount === 0) { | ||
return; | ||
} | ||
} |
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바깥 for문과 안쪽 for문이 있는데, 안쪽 for문에서 한 번도 교환이 일어나지 않았다면, 정렬되어있다는 것을 확신할 수 있습니다.
for (let i = 0; i < array.length; i++) {
let exchangeCount = 0;
for (let j = 0; j < array.length - 1 - i; j++) {
if (array[j] > array[j + 1]) {
exchangeCount += 1;
[array[j + 1], array[j]] = [array[j], array[j + 1]];
}
}
if (exchangeCount === 0) {
return;
}
}
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Choose a reason for hiding this comment
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감사합니다!! 굳이 i === 0을 할 필요가 없었네요!
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우선 구현된 bubble-sort 코드를 참고하지 않고, 혼자 bubble 정렬을 구현해봤습니다.
생각보다 구현하는 것이 어려워서 생각한 내용을 주석으로 적어 놨습니다.
추가 문제랑 다음 과제도 얼른 풀이 올리겠습니다 😃