최주은 / 16주차 #23
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최주은 / 16주차 #23
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dongrri22
reviewed
Apr 6, 2023
| @@ -0,0 +1,47 @@ | |||
| // 인접 행렬을 1차원 배열로 만듦 -> 통과 | |||
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메모리 초과, 시간 초과를 1차원 배열을 이용해 해결할 수 있네요
dongrri22
reviewed
Apr 6, 2023
| let visited = new Array(n); | ||
| for (let i = 0; i < n; i++) visited[i] = new Array(n).fill(false); | ||
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| const dfs = (i, j, height) => { |
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이렇게 각 방향마다 dfs 함수를 호출하는 방법도 있네요
dongrri22
reviewed
Apr 6, 2023
| visited[0] = true; | ||
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| const adjacent = new Array(n + 1); | ||
| // adjacent 배열을 new Array(n + 1).fill(new Array(n + 1).fill(0))로 했을 경우 왜 밑에 forEach문이 제대로 동작하지 않는가? |
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반복문 안에서 자체적으로 new Array( ) 생성자를 반복해주면 어떨까요?
dongrri22
reviewed
Apr 6, 2023
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| const adjacent = new Array(n + 1); | ||
| // adjacent 배열을 new Array(n + 1).fill(new Array(n + 1).fill(0))로 했을 경우 왜 밑에 forEach문이 제대로 동작하지 않는가? | ||
| // 한번에 여러 인덱스가 같이 바뀌어버리는 이유? |
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fill 메서드가 값을 복사해 할당해 주는데, fill의 매개변수로 객체가 전달되어 동일한 주소가 전달되는게 아닐까요?
sena-22
approved these changes
May 5, 2023
| } | ||
| if (cabbageField[nowI]?.[nowJ + 1] && !visited[nowI]?.[nowJ + 1]) { | ||
| visited[nowI][nowJ + 1] = true; | ||
| queue.push([nowI, nowJ + 1]); |
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이 부분은 반복문을 사용하면 더 간결할 것 같아요
| 3. queue 생성 (처음에 V 넣고 시작) | ||
| 4. queue에 있는 노드를 빼고 해당 노드의 (방문하지 않은) 인접 노드들을 queue에 넣는다. 이때 queue에서 빠지면 | ||
| visited 배열에 체크하고, 정답에 추가한다. | ||
| */ |
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