[mintheon] Week 4

merge-two-sorted-lists/mintheon.java

@@ -0,0 +1,36 @@
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode() {}
+ *     ListNode(int val) { this.val = val; }
+ *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
+ * }
+ */
+class Solution {
+  /**
+   시간복잡도: O(N)
+   공간복잡도: O(1)
+   */
+  public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
+    ListNode answer = new ListNode(-1);
+    ListNode node = answer;
+
+    while(list1 != null && list2 != null) {
+      if(list1.val < list2.val) {
+        node.next = list1;
+        list1 = list1.next;
+      } else {
+        node.next = list2;
+        list2 = list2.next;
+      }
+
+      node = node.next;
+    }
+
+    node.next = list1 != null ? list1 : list2;
+
+    return answer.next;
+  }
+}

missing-number/mintheon.java

@@ -0,0 +1,17 @@
+class Solution {
+  /**
+   시간복잡도: O(n)
+   공간복잡도: O(1)
+   */
+  public int missingNumber(int[] nums) {
+    int n = nums.length;
+    int expectedSum = n * (n + 1) / 2;
+
+    int arraySum = 0;
+    for(int i = 0; i < nums.length; i++) {
+      arraySum += nums[i];
+    }
+
+    return expectedSum - arraySum;
+  }
+}

palindromic-substrings/mintheon.java

@@ -0,0 +1,33 @@
+class Solution {
+  /**
+   시간복잡도: O(n^2)
+   공간복잡도: O(1)
+   -> 모든 부분 문자열을 확인하는 방식 대신 좀 더 최적화한 방식으로 다시 풀어볼 것.
+   */
+  public int countSubstrings(String s) {
+    int count = 0;
+
+    for(int i = 0; i < s.length(); i++) {
+      for(int j = i; j < s.length(); j++) {
+        if(isPalindrom(s, i, j)) {
+          count++;
+        }
+      }
+    }
+
+    return count;
+  }
+
+  private boolean isPalindrom(String text, int left, int right) {
+    while(left < right) {
+      if(text.charAt(left) != text.charAt(right)) {
+        return false;
+      }
+
+      left++;
+      right--;
+    }
+
+    return true;
+  }
+}

word-search/mintheon.java

@@ -0,0 +1,45 @@
+class Solution {
+  // BFS로 변환해서 다시 풀어볼 것
+  char[][] board;
+  boolean[][] visited;
+  int[] moveX = {-1, 1, 0, 0};
+  int[] moveY = {0, 0, -1, 1};
+  int row, col;
+
+  public boolean exist(char[][] board, String word) {
+    this.board = board;
+    this.row = board.length;
+    this.col = board[0].length;
+    this.visited = new boolean[row][col];
+
+    for(int y = 0; y < row; y++) {
+      for(int x = 0; x < col; x++) {
+        if(this.hasWord(y, x, word, 0)) {
+          return true;
+        }
+      }
+    }
+
+    return false;
+  }
+
+  private boolean hasWord(int y, int x, String word, int index) {
+    if(index >= word.length()) {
+      return true;
+    }
+
+    if(x < 0 || x >= col || y < 0 || y >= row || visited[y][x] || board[y][x] != word.charAt(index)) {
+      return false;
+    }
+
+    this.visited[y][x] = true;
+    for(int i = 0; i < 4; i++) {
+      if(this.hasWord(y + moveY[i], x + moveX[i], word, index + 1)) {
+        return true;
+      }
+    }
+
+    this.visited[y][x] = false;
+    return false;
+  }
+}