[jj7779607] Week2 #725
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[jj7779607] Week2 #725
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TonyKim9401
reviewed
Dec 21, 2024
Comment on lines
+25
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| while (low < high && nums[low] === nums[low + 1]) low++; | ||
| while (low < high && nums[high] === nums[high - 1]) high--; |
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맞아요 :) 전 문제의 핵심은 정렬, 투 포인터 를 극한의 효율로 사용하기 위한 25,26 라인의 while문인것 같습니다. 전 이 부분을 고려 안해서 자꾸 실패했었네요 ㅎㅎ
TonyKim9401
reviewed
Dec 21, 2024
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| // 규칙성 찾기 | ||
| // n = 1 -> 1 (1) | ||
| // n = 2 -> 2 (1+1, 2) | ||
| // n = 3 -> 2 + 1 = 3 (1+1+1, 1+2, 2+1) | ||
| // n = 4 -> 2 + 3 = 5 (1+1+1, 1+1+2, 1+2+1, 2+1+1, 2+2) | ||
| // n = 5 -> 3 + 5 = 8 (1+1+1+1+1, 1+1+1+2, 1+1+2+1, 1+2+1+1, 2+1+1+1, 1+2+2, 2+2+1, 2+1+2) |
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규칙성을 잘 찾으셔서 무리없이 풀으셨을것 같습니다!
이런 식의 규칙성 문제는 대부분 DP를 사용하는것 같아요!
TonyKim9401
reviewed
Dec 21, 2024
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| for (let i = 0; i < s.length; i++) { | ||
| countS[s[i]] = (countS[s[i]] || 0) + 1; | ||
| countT[t[i]] = (countT[t[i]] || 0) + 1; | ||
| } | ||
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| // 두 객체 동일하면 true, 아니면 false | ||
| for (let char in countS) { | ||
| if (countS[char] !== countT[char]) { | ||
| return false; | ||
| } | ||
| } |
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같은 방법이긴 합니다만 count배열을 하나만 두고 같은 값이 s에 있으면 +, t에 있으면 -를 해주면 결국 해당 값이 0이 될거에요.
이때 두번째 for문에서 배열의 값이 모두 0이면 anagram이 성립하는 식으로 풀이 방법도 있습니다 :)
TonyKim9401
approved these changes
Dec 21, 2024
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