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[shinsj4653] Week 14 Solutions #1634
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""" | ||
[문제풀이] | ||
# Inputs | ||
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# Outputs | ||
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# Constraints | ||
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# Ideas | ||
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[회고] | ||
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""" | ||
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""" | ||
[문제풀이] | ||
# Inputs | ||
n: int | ||
# Outputs | ||
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ans: arr | ||
길이: n + 1 | ||
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ans내 각 원소들 : ans[i]: i의 이진법에서 1의 개수 | ||
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# Constraints | ||
0 <= n <= 10^5 | ||
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# Ideas | ||
2중 for문 이면 안됨 | ||
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[회고] | ||
dp를 활용한 풀이도 같이 알아두자 | ||
""" | ||
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# 내 풀이 | ||
class Solution: | ||
def countBits(self, n: int) -> List[int]: | ||
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ans = [] | ||
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for i in range(n + 1): | ||
if i == 0 or i == 1: | ||
ans.append(i) | ||
continue | ||
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num = i | ||
cnt = 0 | ||
while num > 0: | ||
num, n = num // 2, num % 2 | ||
if n == 1: | ||
cnt += 1 | ||
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ans.append(cnt) | ||
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return ans | ||
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# 해설 | ||
class Solution: | ||
def countBits(self, n: int) -> List[int]: | ||
dp = [0] * (n + 1) | ||
for num in range(1, n + 1): | ||
dp[num] = dp[num // 2] + (num % 2) | ||
return dp | ||
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""" | ||
[문제풀이] | ||
# Inputs | ||
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# Outputs | ||
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# Constraints | ||
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# Ideas | ||
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[회고] | ||
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""" | ||
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""" | ||
[문제풀이] | ||
# Inputs | ||
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# Outputs | ||
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# Constraints | ||
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# Ideas | ||
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[회고] | ||
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""" | ||
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""" | ||
[문제풀이] | ||
# Inputs | ||
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# Outputs | ||
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# Constraints | ||
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# Ideas | ||
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[회고] | ||
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""" | ||
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가장 오른쪽 비트를 기준으로 하는 DP로 풀이하신 것 같네요!
저의 경우에는
// 2
나% 2
연산은 비트 연산으로 대체하는 편인데요, 비트 연산이 성능 측면에서 미세하게나마 더 효율적이라고 합니다! 따라서dp[num] = dp[num >> 1] + (num & 1)
로도 작성할 수 있을 것 같아요.또한, 가장 왼쪽 비트를 기준으로 하는 DP로도 한 번 풀이해보시는 걸 추천드려요~!