Merge pull request #916 from eunhwa99/main

[eunhwa99] week 7

Author: eunhwa99

longest-substring-without-repeating-characters/eunhwa99.java

@@ -0,0 +1,29 @@
+import java.util.HashMap;
+import java.util.Map;
+
+class Solution {
+  // 시간 복잡도: O(N)
+  // 공간 복잡도: O(N)
+  public int lengthOfLongestSubstring(String s) {
+    Map<Character, Integer> position = new HashMap<>();
+    int start = 0; // substring의 시작점
+    int maxLength = 0;
+
+    for (int idx = 0; idx < s.length(); idx++) {
+      char currentChar = s.charAt(idx);
+
+      // 같은 문자가 이미 map 에 있고, 그 문자가 현재 substring에 포함된 문자인지 확인
+      if (position.containsKey(currentChar) && position.get(currentChar) >= start) {
+        start = position.get(currentChar) + 1;
+        // 같은 문자가 포함되지 않게 substring의 시작을 옮긴다.
+      }
+
+      maxLength = Math.max(maxLength, idx - start + 1);
+
+      position.put(currentChar, idx);
+    }
+
+    return maxLength;
+  }
+}
+

number-of-islands/eunhwa99.java

@@ -0,0 +1,67 @@
+import java.util.LinkedList;
+import java.util.Queue;
+
+// BFS 사용
+// 시간 복잡도 : O(MxN)
+// 공간 복잡도: O(MxN)
+class Solution {
+
+  int[][] dir = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
+
+  public int numIslands(char[][] grid) {
+    int row = grid.length;
+    int col = grid[0].length;
+
+    int total = 0;
+    for (int i = 0; i < row; i++) {
+      for (int j = 0; j < col; j++) {
+        if (grid[i][j] == '1') {
+          total++;
+          BFS(grid, i, j, row, col);
+          System.out.println(grid[i][j]);
+        }
+      }
+    }
+    return total;
+  }
+
+  private void BFS(char[][] grid, int r, int c, int sizeR, int sizeC) {
+    Queue<Position> queue = new LinkedList<>();
+
+    queue.add(new Position(r, c));
+    grid[r][c] = '0'; // '0'으로 변경 (방문 체크)
+
+    while (!queue.isEmpty()) {
+      Position current = queue.poll();
+      int curR = current.r;
+      int curC = current.c;
+
+      for (int i = 0; i < 4; i++) {
+        int dirR = dir[i][0];
+        int dirC = dir[i][1];
+
+        int nextR = curR + dirR;
+        int nextC = curC + dirC;
+
+        if (nextR < 0 || nextR >= sizeR || nextC < 0 || nextC >= sizeC || grid[nextR][nextC] == '0') {
+          continue;
+        }
+        queue.add(new Position(nextR, nextC));
+        grid[nextR][nextC] = '0';
+      }
+    }
+
+  }
+
+  static class Position {
+
+    int r;
+    int c;
+
+    Position(int r, int c) {
+      this.r = r;
+      this.c = c;
+    }
+  }
+}
+

reverse-linked-list/eunhwa99.java

@@ -0,0 +1,29 @@
+/**
+ * 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; }
+ * }
+ */
+
+// 시간 복잡도: O(N)
+// 공간복잡도: O(N)
+class Solution {
+  public ListNode reverseList(ListNode head) {
+    if(head==null) return head;
+
+    ListNode pointer = new ListNode(head.val);
+
+    ListNode tempPointer;
+    while(head.next!=null){
+      tempPointer = new ListNode(head.next.val, pointer);
+      pointer = tempPointer;
+      head = head.next;
+    }
+  }
+    return pointer;
+}
+

set-matrix-zeroes/eunhwa99.java

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+import java.util.HashSet;
+import java.util.Set;
+
+// 시간 복잡도: O(row x col)
+// 공간 복잡도: O(row + col)
+class Solution {
+
+  public void setZeroes(int[][] matrix) {
+    int row = matrix.length;
+    int col = matrix[0].length;
+
+    // 행과 열에 0이 있는지 체크할 배열
+    Set<Integer> rowZero = new HashSet<>();
+    Set<Integer> colZero = new HashSet<>();
+
+    for (int i = 0; i < row; i++) {
+      for (int j = 0; j < col; j++) {
+        if (matrix[i][j] == 0) {
+          rowZero.add(i);
+          colZero.add(j);
+        }
+      }
+    }
+
+    // 행을 0으로 설정
+    for (int r : rowZero) {
+      for (int c = 0; c < col; c++) {
+        matrix[r][c] = 0;
+      }
+    }
+
+    // 열을 0으로 설정
+    for (int c : colZero) {
+      for (int r = 0; r < row; r++) {
+        matrix[r][c] = 0;
+      }
+    }
+  }
+}
+
+

unique-paths/eunhwa99.java

@@ -0,0 +1,23 @@
+class Solution {
+
+  public int uniquePaths(int m, int n) {
+    int[][] paths = new int[m][n];
+
+    for (int i = 0; i < m; i++) {
+      paths[i][0] = 1; //가장 왼쪽 줄은 항상 경로가 1개
+    }
+
+    for (int i = 0; i < n; i++) {
+      paths[0][i] = 1; // 가장 윗줄은 항상 경로가 1개
+    }
+    for (int i = 1; i < m; i++) {
+      for (int j = 1; j < n; j++) {
+
+        paths[i][j] = paths[i - 1][j] + paths[i][j - 1];
+      }
+    }
+
+    return paths[m - 1][n - 1];
+  }
+}
+