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Commit b9ee027

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[20220826] leetcode study-plan 문제 풀이
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# https://leetcode.com/problems/is-subsequence/
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# 392. Is Subsequence
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class Solution:
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def isSubsequence(self, s: str, t: str) -> bool:
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i, j = 0, 0
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while i < len(s) and j < len(t):
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if s[i] == t[j]:
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i += 1
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j += 1
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else:
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j += 1
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return i == len(s)
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if __name__ == '__main__':
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sol = Solution()
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print(sol.isSubsequence("abc", "ahbgdc"))
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print(sol.isSubsequence("axc", "ahbgdc"))
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print(sol.isSubsequence("b", "abc"))
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# https://leetcode.com/problems/isomorphic-strings/
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# 205. Isomorphic Strings
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from collections import defaultdict
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class Solution:
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def isIsomorphic(self, s: str, t: str) -> bool:
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sdic = defaultdict()
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tdic = defaultdict()
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if len(s) != len(t):
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return False
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for a, b in zip(s, t):
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sdic[a] = b
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tdic[b] = a
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for a, b in zip(s, t):
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if sdic[a] != b or tdic[b] != a:
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return False
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return True
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if __name__ == '__main__':
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sol = Solution()
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print(sol.isIsomorphic("egg", "add"))
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print(sol.isIsomorphic("foo", "bar"))
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print(sol.isIsomorphic("paper", "title"))
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print(sol.isIsomorphic("badc", "baba"))
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# https://leetcode.com/problems/merge-two-sorted-lists/
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# 21. Merge Two Sorted Lists
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from typing import Optional
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# Definition for singly-linked list.
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class ListNode:
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def __init__(self, val=0, next=None):
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self.val = val
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self.next = next
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class Solution:
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def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
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# ListNode 생성 하여 풀이 (AC)
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if list1 is not None and list2 is None:
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return list1
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if list1 is None and list2 is not None:
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return list2
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if list1 is None and list2 is None:
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return None
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start = dummy = ListNode()
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while list1 is not None and list2 is not None:
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if list1.val < list2.val:
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start.next = list1
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list1 = list1.next
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elif list1.val == list2.val:
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start.next = list1
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list1 = list1.next
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else:
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start.next = list2
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list2 = list2.next
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start = start.next
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if list1 is not None and list2 is None:
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start.next = list1
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if list2 is not None and list1 is None:
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start.next = list2
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return dummy.next
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def mergeTwoLists_2(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
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# ListNode 생성 없이 풀이 (Discuss 참고)
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# 작은 값을 가진 리스트 노드를 이동 시켜서 연결 시키면 됨
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# list1이 None 이면 list2 가 list1의 가장 큰 값 보다 크다는 의미이므로
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# list2 를 리턴시킨다.
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if list1 is None:
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return list2
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if list2 is None:
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return list1
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if list1.val < list2.val:
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# list1을 이동시킴
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# 그리고 결과로 나온 값 (최소 l1 보다 큰 값) 을 l1 에 이어 붙인다.
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list1.next = self.mergeTwoLists_2(list1.next, list2)
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return list1
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else:
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# list2 이동
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# 같으면 사실 list1 이든 list2 든 누가 이동해도 상관 없음 (어느쪽에 붙여든 상관 없음)
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list2.next = self.mergeTwoLists_2(list1, list2.next)
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return list2
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# https://leetcode.com/problems/reverse-linked-list/
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# 206. Reverse Linked List
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from typing import Optional
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# Definition for singly-linked list.
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class ListNode:
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def __init__(self, val=0, next=None):
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self.val = val
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self.next = next
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def __str__(self) -> str:
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return str(self.val)
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class Solution:
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def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
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if head is None:
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return None
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start = dummy = ListNode()
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stack = []
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while head is not None:
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stack.append(head.val)
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head = head.next
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while stack:
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start.next = ListNode(stack.pop())
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start = start.next
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return dummy.next
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def reverseList_2(self, head: Optional[ListNode]) -> Optional[ListNode]:
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# TODO stack 사용 없이 풀어보기
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pass
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if __name__ == '__main__':
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sol = Solution()
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cur = node = ListNode()
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for i in range(1, 6):
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node.next = ListNode(i)
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node = node.next
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print(sol.reverseList_2(cur.next))

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