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191. Number of 1 Bits class Solution: def hammingWeight(self, n: int) -> int: res = 0 while n: # n > 0: res += n % 2 # res += n&1 n = n >> 1 # n = n // 2 return res https://youtu.be/5Km3utixwZs?si=f7G0eyMnnaO_-VMR class Solution: def hammingWeight(self, n: int) -> int: res = 0 while n: # n > 0: n &= (n-1) # n = n & ..

190. Reverse Bits https://youtu.be/UcoN6UjAI64?si=QcqwmCh09NAZbJMq class Solution: def reverseBits(self, n: int) -> int: # 1과 & 오퍼레이션하면, 해당 비트 나옴 # > i): n을 i만큼 오른쪽으로 쉬프트하면 i번째 비트가 오른쪽 끝에 위치함 # & 1: AND 연산을 통해 해당 비트를 추출 (나머지는 0) bit = (n >> i) & 1 # 추출한 비트를 뒤집은 위치로 이동 (왼쪽에서부터 채우기) # 31-i: 0번째 비트는 31번째에, 1번째 비트는 30번째에 위치하게 됨 ..

67. Add Binary https://youtu.be/keuWJ47xG8g?si=9FQTur6Jph3SV28p class Solution: def addBinary(self, a: str, b: str) -> str: res = "" carry = 0 a, b = a[::-1], b[::-1] for i in range(max(len(a), len(b))): digitA = int(a[i]) if i

210. Course Schedule II Hint 1 This problem is equivalent to finding the topological order in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.  Hint 2 Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort. Hint 3 Topological sort could also be done vi..