[Chaedie] Week 14

binary-tree-level-order-traversal/Chaedie.py

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+"""
+Solution: BFS
+Time: O(n)
+Space: O(n)
+"""
+
+class Solution:
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:        
+        if not root: return []
+
+        q = deque([root])
+        result = []
+        while q:
+            level = []
+            for i in range(len(q)):
+                node = q.popleft()
+                level.append(node.val)
+                if node.left: q.append(node.left)
+                if node.right: q.append(node.right)
+            result.append(level)
+
+        return result

counting-bits/Chaedie.py

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+
+"""
+Solution: 
+    1) bin() 메서드로 binary 만들어주고 
+    2) 1 의 갯수를 세어준다.
+Time: O(n^2)
+Space: O(1)
+"""
+
+class Solution:
+    def countBits(self, n: int) -> List[int]:
+        result = [0 for i in range(n + 1)]
+
+        for i in range(n+1):
+            b = bin(i)
+            count = 0
+            for char in b:
+                if char == '1':
+                    count += 1
+            result[i] = count
+
+        return result
+
+"""
+Solution: 
+    1) 2로 나눈 나머지가 1bit 이라는 성질을 이용해서 count
+Time: O(n logn)
+Space: O(1)
+"""
+
+class Solution:
+    def countBits(self, n: int) -> List[int]:
+        def count(num):
+            count = 0
+            while num > 0:
+                count += num % 2 
+                num = num // 2
+            return count
+
+        return [count(i) for i in range(n+1)]
+
+"""
+Solution:
+    1) LSB 가 0 1 0 1 반복되므로 num % 2 를 사용한다.
+    2) 나머지 빗은 LSB를 제외한 값이므로 num // 2 를 사용한다.
+Time: O(n)
+Space: O(1)
+"""
+
+class Solution:
+
+    def countBits(self, n: int) -> List[int]:
+        dp = [0 for i in range(n+1)]
+
+        for i in range(1, n+1):
+            LSB = i % 2
+            dp[i] = dp[i // 2] + LSB
+
+        return dp