Merge pull request #539 from kjb512/main

[kayden] Week 10 Solutions

Author: ijlijijij

course-schedule/kayden.py

@@ -0,0 +1,33 @@
+from collections import deque
+class Solution:
+    # 시간복잡도: O(numCourses + prerequisites의 길이)
+    # 공간복잡도: O(numCourses + prerequisites의 길이)
+    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
+
+        reachable = [0 for _ in range(numCourses)]
+        graph = [[] for _ in range(numCourses)]
+
+        for a, b in prerequisites:
+            reachable[a] += 1
+            graph[b].append(a)
+
+        q = deque()
+        visited = set()
+        for i in range(numCourses):
+            if reachable[i] == 0:
+                q.append(i)
+                visited.add(i)
+
+        while q:
+            node = q.popleft()
+
+            for next_node in graph[node]:
+                reachable[next_node] -= 1
+                if next_node not in visited and reachable[next_node] == 0:
+                    q.append(next_node)
+                    visited.add(next_node)
+
+        if len(visited) == numCourses:
+            return True
+
+        return False

invert-binary-tree/kayden.py

@@ -0,0 +1,19 @@
+class Solution:
+    # 시간복잡도: O(N)
+    # 공간복잡도: O(1)
+    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
+        def invert(node):
+            if not node:
+                return
+
+            r = invert(node.left)
+            l = invert(node.right)
+
+            node.right = r
+            node.left = l
+
+            return node
+
+        invert(root)
+
+        return root

jump-game/kayden.py

@@ -0,0 +1,15 @@
+class Solution:
+    # 시간복잡도: O(N)
+    # 공간복잡도: O(1)
+    # 이 문제는 마지막 테케가 통과하지 않아서 답을 참고했습니다.
+    def canJump(self, nums: List[int]) -> bool:
+
+        gas = 0
+        for n in nums:
+            if gas < 0:
+                return False
+            elif n > gas:
+                gas = n
+            gas -= 1
+
+        return True

merge-k-sorted-lists/kayden.py

@@ -0,0 +1,41 @@
+class Solution:
+    # 시간복잡도: O(NlogK) N: 모든 리스트의 노드 수 합 K: 리스트의 개수
+    # 공간복잡도: O(1) 기존 노드 재활용
+    def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:
+        if not lists: return None
+        if len(lists) == 1: return lists[0]
+
+        def merge(a, b):
+            res = ListNode()
+            cur = res
+
+            while a and b:
+                if a.val > b.val:
+                    cur.next = b
+                    b = b.next
+                else:
+                    cur.next = a
+                    a = a.next
+                cur = cur.next
+
+            if a:
+                cur.next = a
+            else:
+                cur.next = b
+
+            return res.next
+
+        def mergeK(lo, hi):
+            if lo == hi:
+                return lists[lo]
+
+            if hi - lo == 1:
+                return merge(lists[lo], lists[hi])
+
+            mid = (lo + hi) // 2
+            left = mergeK(lo, mid)
+            right = mergeK(mid+1, hi)
+
+            return merge(left, right)
+
+        return mergeK(0, len(lists)-1)

search-in-rotated-sorted-array/kayden.py

@@ -0,0 +1,22 @@
+class Solution:
+    # 시간복잡도: O(logN)
+    # 공간복잡도: O(1)
+    def search(self, nums: List[int], target: int) -> int:
+        n = len(nums)
+        st, en = 0, n-1
+        while st <= en:
+            mid = (st+en)//2
+            if nums[mid] == target:
+                return mid
+            elif nums[mid] >= nums[st]:
+                if nums[st] <= target <= nums[mid]:
+                    en = mid - 1
+                else:
+                    st = mid + 1
+            else:
+                if nums[mid] <= target <= nums[en]:
+                    st = mid + 1
+                else:
+                    en = mid - 1
+
+        return -1