[hyeri0903] WEEK 08 Solutions

clone-graph/hyeri0903.java

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+/*
+// Definition for a Node.
+class Node {
+    public int val;
+    public List<Node> neighbors;
+    public Node() {
+        val = 0;
+        neighbors = new ArrayList<Node>();
+    }
+    public Node(int _val) {
+        val = _val;
+        neighbors = new ArrayList<Node>();
+    }
+    public Node(int _val, ArrayList<Node> _neighbors) {
+        val = _val;
+        neighbors = _neighbors;
+    }
+}
+*/
+
+class Solution {
+    //key: original node, value: clone node
+    private Map<Node, Node> map = new HashMap<>();
+
+    public Node cloneGraph(Node node) {
+        if(node == null) {
+            return null;
+        }
+        //이미 존재하면 return
+        if(map.containsKey(node)) {
+            return map.get(node);
+        }
+
+        //현재 노드 복제
+        Node clone = new Node(node.val);
+        map.put(node, clone);
+
+        //이웃 노드들 복제해서 연결
+        for(Node neighbor: node.neighbors) {
+            clone.neighbors.add(cloneGraph(neighbor));
+        }
+        return clone;
+    }
+}

longest-common-subsequence/hyeri0903.java

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+class Solution {
+    public int longestCommonSubsequence(String text1, String text2) {
+        /**
+        1.문제: 가장 긴 common subsequence length return, 없으면 0 return
+        2.constraints
+        - length min = 1, max = 1000
+        - text1, text2 모두 lowercase로 구성
+        3.풀이
+        - dp[i][j] = text1의 앞 i개, text2의 앞 j개를 비교했을 때 LCS 길이
+        - 문자가 같으면: dp[i][j] = dp[i-1][j-1] + 1
+        - 문자가 다르면: dp[i][j] = max(dp[i-1][j], dp[i][j-1]) 
+         */
+        int t1Size = text1.length(); //row
+        int t2Size = text2.length(); //col
+        int[][] dp = new int[t1Size + 1][t2Size + 1];
+
+        for(int i = 1; i <= t1Size; i++) {
+            for(int j = 1; j <= t2Size; j++) {
+                if(text1.charAt(i-1) == text2.charAt(j-1)) {
+                    dp[i][j] = dp[i-1][j-1] + 1;
+                } else {
+                    dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]);
+                }
+            }
+        }
+        return dp[t1Size][t2Size];
+    }
+}

longest-repeating-character-replacement/hyeri0903.java

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+class Solution {
+    public int characterReplacement(String s, int k) {
+        /**
+        1.problem: k번 특정 문자를 골라서 바꿀 수 있다. 이때 same letter 로 이루어진 가장 긴 substring 의 길이을 구하라.
+        2.constraints
+        - s.length min = 1, max = 10^5
+        - k.length min = 0, max = s.length
+        3.solution
+        - sliding window: time O(n), space O(1)
+         */
+
+        int[] table = new int[26];
+        int n = s.length();
+        int maxFreq = 0;
+        int maxLen = 0;
+        int left = 0;
+
+        for(int right = 0; right < n; right++) {
+            //1.right 확장 -> count 증가
+            int currentChar = s.charAt(right) - 'A';
+            table[currentChar]++;
+
+            //2.maxFreq update
+            maxFreq = Math.max(maxFreq, table[currentChar]);
+
+            //3.조건 깨지면 left 이동하면서 count--
+            //바꿔야하는 문자 개수가 k 보다 큰 경우
+            if((right - left + 1 - maxFreq) > k) {
+                int idx = s.charAt(left) - 'A';
+                table[idx]--;
+                left++;
+            }
+
+            //4.maxLen update
+            maxLen = Math.max(maxLen, right - left + 1);
+        }
+        return maxLen;
+    }
+}

palindromic-substrings/hyeri0903.java

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+class Solution {
+    public int countSubstrings(String s) {
+         /**
+        1.문제: s 에서 palindromoic substring 개수 return
+        2.constraints:
+        - 연속적인 문자
+        - s.length min = 1, max = 1000
+        3.풀이
+        -  bruteforce: time:O(n^3), space:O(n)
+        - expand around : time(On^2), space:O(1)
+         */
+
+         int n = s.length();
+         if(n == 1) return 1;
+        int count = 0;
+
+        for(int i = 0; i < n; i++) {
+            //odd
+            count += checkPalindrome(s, i, i);
+
+            //even
+            count += checkPalindrome(s, i, i+1);
+        }
+        return count;
+
+    }
+
+    private int checkPalindrome(String s, int left, int right) {
+        int count = 0;
+        while(left >= 0 && right < s.length() && s.charAt(right) == s.charAt(left)) {
+            count += 1;
+            left -= 1;
+            right += 1;
+        }
+        return count;
+    }
+
+}

reverse-bits/hyeri0903.java

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+class Solution {
+    public int reverseBits(int n) {
+        int result = 0;
+        for(int i = 0; i < 32; i++) {
+            result <<= 1;   //왼쪽 이동
+            result |= (n & 1); //마지막 비트 붙이기
+            n >>= 1;    // 오른쪽 이동
+        }
+        return result;
+    }
+}