[GangBean] Week 2 #707
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[GangBean] Week 2 #707
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TonyKim9401
approved these changes
Dec 20, 2024
| - space: O(1) -> no extra space is needed | ||
| */ | ||
| // 0. assign return variable Set | ||
| Set<List<Integer>> answer = new HashSet<>(); |
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중복을 피하려고 Set을 사용한게 인상깊네요..!
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| if (preorder.length == 0) return null; | ||
| if (preorder.length == 1) return new TreeNode(preorder[0]); | ||
| int i = 0; | ||
| List<Integer> leftPreorder = new ArrayList<>(); // O(N) | ||
| List<Integer> leftInorder = new ArrayList<>(); // O(N) | ||
| List<Integer> rightPreorder = new ArrayList<>(); // O(N) | ||
| List<Integer> rightInorder = new ArrayList<>(); // O(N) | ||
| for (; i < inorder.length; i++) { // O(N) | ||
| if (inorder[i] == preorder[0]) break; | ||
| leftPreorder.add(preorder[i+1]); | ||
| leftInorder.add(inorder[i]); | ||
| } | ||
| for (int idx = i+1; idx < inorder.length; idx++) { // O(N) | ||
| rightPreorder.add(preorder[idx]); | ||
| rightInorder.add(inorder[idx]); | ||
| } | ||
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| return new TreeNode(preorder[0], buildTree(leftPreorder.stream().mapToInt(Integer::intValue).toArray(), leftInorder.stream().mapToInt(Integer::intValue).toArray()), buildTree(rightPreorder.stream().mapToInt(Integer::intValue).toArray(), rightInorder.stream().mapToInt(Integer::intValue).toArray())); |
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와; 이렇게 푸시는건 상상을 못했네요. dfs로 풀어보시는 것도 좋을것 같습니다!
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| Map<Character, Integer> sMap = new HashMap<>(); | ||
| Map<Character, Integer> tMap = new HashMap<>(); | ||
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| for (char c: s.toCharArray()) { | ||
| sMap.put(c, sMap.getOrDefault(c, 0) + 1); | ||
| } | ||
| for (char c: t.toCharArray()) { | ||
| tMap.put(c, tMap.getOrDefault(c, 0) + 1); | ||
| } | ||
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| return Objects.equals(sMap, tMap); |
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전 s와 t를 .toCharArray()로 바꾼 후 Arrays.sort -> Arrays.equals() 로 비교했는데 Objects.equals로 HashMap 자체를 비교하는 방법도 있군요. 배워갑니다!
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