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[아현] WEEK02 Solutions #349
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Original file line number | Diff line number | Diff line change |
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/** | ||
* Definition for a binary tree node. | ||
* public class TreeNode { | ||
* int val; | ||
* TreeNode left; | ||
* TreeNode right; | ||
* TreeNode() {} | ||
* TreeNode(int val) { this.val = val; } | ||
* TreeNode(int val, TreeNode left, TreeNode right) { | ||
* this.val = val; | ||
* this.left = left; | ||
* this.right = right; | ||
* } | ||
* } | ||
*/ | ||
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// time : O(n) | ||
// space : O(n) | ||
// n은 트리 노드 수 | ||
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class Solution { | ||
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private int i = 0; | ||
Map<Integer, Integer> map = new HashMap<>(); | ||
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public TreeNode buildTree(int[] preorder, int[] inorder) { | ||
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for(int i = 0; i < inorder.length; i++) { | ||
map.put(inorder[i], i); | ||
} | ||
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return build(preorder, inorder, 0, inorder.length); | ||
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} | ||
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private TreeNode build(int[] preorder, int[] inorder, int start, int end) { | ||
if(i >= preorder.length || start >= end) { | ||
return null; | ||
} | ||
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int value = preorder[i++]; | ||
int index = map.get(value); | ||
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TreeNode leftTreeNode = build(preorder, inorder, start, index); | ||
TreeNode rightTreeNode = build(preorder, inorder, index+1, end); | ||
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return new TreeNode(value, leftTreeNode, rightTreeNode); | ||
} | ||
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} |
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//time : O(n) | ||
//space : O(n) | ||
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class Solution { | ||
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// f(n) = f(n >> 1) + (n & 1) | ||
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public int[] countBits(int n) { | ||
int[] answer = new int[n + 1]; | ||
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answer[0] = 0; | ||
for(int i = 1; i <= n; i++) { | ||
answer[i] = answer[i >> 1] + (i & 1); | ||
} | ||
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return answer; | ||
} | ||
} |
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. charArray가 반복문을 위해서만 사용된 것 같아서 추가적으로 n 이 무엇을 의미하는 걸까요? There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. n 은 문자열의 길이를 나타냅니다. 말씀하신 데로 charAt을 쓰면 sToChar와 tToChar 를 사용하지 않아도 되어서 공간복잡도가 O(1)로 끝날 수 있어서 좋은 선택지인 것 같습니다! 👍 |
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// time : O(n) | ||
// space : O(1) | ||
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class Solution { | ||
public boolean isAnagram(String s, String t) { | ||
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if(s.length() != t.length()) return false; | ||
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int[] charCount = new int[26]; | ||
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for(int i = 0; i < s.length(); i++) { | ||
charCount[s.charAt(i) - 'a']++; | ||
charCount[t.charAt(i) - 'a']--; | ||
} | ||
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for(int i : charCount) { | ||
if(i != 0) { | ||
return false; | ||
} | ||
} | ||
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return true; | ||
} | ||
} | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 나머지 문제도 시간 되실 때 꼭 풀어보시면 좋을것 같습니다!! 배울게 많은 문제들이었던 것 같습니다=) |
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The reason will be displayed to describe this comment to others. Learn more.
nit: 접근 제한자를 일관적으로 붙이거나 생략해주면 불필요한 오해의 소지를 없앨 수 있을 것 같아요.