diff --git a/longest-substring-without-repeating-characters/hyeri0903.java b/longest-substring-without-repeating-characters/hyeri0903.java
new file mode 100644
index 0000000000..ecf404f823
--- /dev/null
+++ b/longest-substring-without-repeating-characters/hyeri0903.java
@@ -0,0 +1,28 @@
+class Solution {
+    public int lengthOfLongestSubstring(String s) {
+        /**
+        1.prob: 중복없는 가장 긴 substring length return
+        2.constraints:
+        - alphabet, digit, space 로 구성
+        - s.length min = 0, max = 50,000
+        3.solution
+        - slding window, time: O(n), space: O(n) or O(1)
+         */
+
+         int maxLen = 0;
+         int left = 0;
+         Set<Character> visited = new HashSet<>();
+
+         for(int i = 0; i < s.length(); i++){
+            char ch = s.charAt(i);
+            while(visited.contains(ch)) {
+                visited.remove(s.charAt(left));
+                left += 1;
+            }
+            visited.add(ch);
+            maxLen = Math.max(maxLen, i - left + 1);
+         }
+    
+        return maxLen;
+    }
+}
diff --git a/number-of-islands/hyeri0903.java b/number-of-islands/hyeri0903.java
new file mode 100644
index 0000000000..bb24a3a98e
--- /dev/null
+++ b/number-of-islands/hyeri0903.java
@@ -0,0 +1,43 @@
+class Solution {
+
+    public int numIslands(char[][] grid) {
+        /**
+        1.prob:섬 개수 구하기
+        2.constraints: 
+        - 원소값 0 or 1
+        - m,n 길이는 최소 = 1, 최대 = 300
+        3.solution: dfs
+         */
+
+        //m: 세로, n: 가로 길이
+        int m = grid.length;
+        int n = grid[0].length;
+        int count = 0;
+
+        for(int i = 0; i < m; i++) {
+            for(int j = 0; j < n; j++) {
+                if(grid[i][j] == '1' && grid[i][j] != '#') {
+                    dfs(i, j, grid, m, n);
+                    count += 1;
+                }
+            }
+        }
+        return count;
+    }
+     void dfs(int i, int j, char[][] grid, int m, int n) {
+            if(i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == '0') {
+                return;
+            }
+            //이미 visited
+            if(grid[i][j] == '#') {
+                return;
+            }
+            //방문 체크
+            grid[i][j] = '#';
+
+            dfs(i+1, j, grid, m, n);
+            dfs(i-1, j, grid, m, n);
+            dfs(i, j+1, grid, m, n);
+            dfs(i, j-1, grid, m, n);
+        }
+}
diff --git a/number-of-islands/hyeri0903.py b/number-of-islands/hyeri0903.py
index be60281602..36a8bc34f2 100644
--- a/number-of-islands/hyeri0903.py
+++ b/number-of-islands/hyeri0903.py
@@ -6,7 +6,6 @@ def numIslands(self, grid: List[List[str]]) -> int:
         '''
         m = len(grid)
         n = len(grid[0])
-        visited = [ [0] * n for _ in range(m)]
         count = 0
 
         def dfs(i, j):
@@ -14,10 +13,10 @@ def dfs(i, j):
             if i < 0 or i >= m or j < 0 or j >= n or grid[i][j] == "0":
                 return
 
-            if visited[i][j] == 1:
+            if grid[i][j] == "#":
                 return
-
-            visited[i][j] = 1 #방문표시
+            #방문표시
+            grid[i][j] = "#"
 
             dfs(i+1, j)
             dfs(i-1, j)
@@ -26,7 +25,7 @@ def dfs(i, j):
         
         for i in range(m):
             for j in range(n):
-                if grid[i][j] == "1" and visited[i][j] == 0:
+                if grid[i][j] == "1" and grid[i][j] != "#":
                     dfs(i,j)
                     count += 1
         return count
diff --git a/reverse-linked-list/hyeri0903.java b/reverse-linked-list/hyeri0903.java
new file mode 100644
index 0000000000..456b6ec8a7
--- /dev/null
+++ b/reverse-linked-list/hyeri0903.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 {
+    public ListNode reverseList(ListNode head) {
+        if(head == null) {
+            return null;
+        }
+
+        Deque<Integer> stack = new ArrayDeque<>();
+        ListNode node = head;
+
+        while(node != null) {
+            stack.push(node.val);
+            node = node.next;
+        }
+
+        ListNode dummy = new ListNode();
+        ListNode cur = dummy;
+
+        while(stack.size() > 0) {
+            int val = stack.pop();
+            cur.next = new ListNode(val);
+            cur = cur.next;
+        }   
+
+        return dummy.next;
+    }
+}
diff --git a/set-matrix-zeroes/hyeri0903.java b/set-matrix-zeroes/hyeri0903.java
new file mode 100644
index 0000000000..4a4697203e
--- /dev/null
+++ b/set-matrix-zeroes/hyeri0903.java
@@ -0,0 +1,37 @@
+class Solution {
+    public void setZeroes(int[][] matrix) {
+        /**
+        1.문제: 0이 존재하는 위치의 모든 row, column을 0으로 set
+        2.constraints: 
+        - m,n min = 1, max = 200
+        - space: O(mn)으로 풀이하지말 것,
+        3.solution
+        - 0의 위치를 확인 -> 0의 위치는 여러개일 수 있음
+        - row, col 각각 0의 위치 저장 
+        - time: O(mn), space O(m+n)
+         */
+        int m = matrix.length;
+        int n = matrix[0].length;
+        int[] row = new int[m]; //0이 존재하는 row 위치이면 1, 아니면 0
+        int[] col = new int[n]; //0이 존재하는 col 위치이면 1, 아니면 0
+
+        int x = 0; int y = 0; //0의 위치
+
+        for(int i = 0; i < m; i++) {
+            for(int j = 0; j < n; j++) {
+                if(matrix[i][j] == 0) {
+                    row[i] = 1;
+                    col[j] = 1;
+                }
+            }
+        }
+
+        for(int i = 0; i < m; i++) {
+            for(int j = 0; j < n; j++) {
+                if(row[i] == 1 || col[j] == 1) {
+                    matrix[i][j] = 0;
+                }
+            }
+        }
+    }
+}
diff --git a/unique-paths/hyeri0903.java b/unique-paths/hyeri0903.java
new file mode 100644
index 0000000000..df0ecf942c
--- /dev/null
+++ b/unique-paths/hyeri0903.java
@@ -0,0 +1,33 @@
+class Solution {
+    public int uniquePaths(int m, int n) {
+        /**
+        1.문제: finish로 가는 unique path 수
+        2.constraints:
+        - m: 세로, n = 가로
+        - m,n min = 1, max = 100
+        - right, down 으로만 움직이기 가능
+        3.solution
+        - dfs ?? => x, 모든 경로 탐색하지 않음
+        - dp => [i][j] = [i-1][j] + [i][j-1]
+        - time: O(mn), space: O(n)
+         */
+
+         int[][] dp = new int[m][n];
+
+        //첫행, 첫열 = 1
+         for(int i =0; i<m; i++) {
+            dp[i][0] = 1;
+         }
+
+         for(int i = 0; i < n; i++) {
+            dp[0][i] = 1;
+         }
+
+         for(int i = 1; i< m; i++) {
+            for(int j = 1; j < n; j++) {
+                dp[i][j] = dp[i-1][j] + dp[i][j-1];
+            }
+         }
+        return dp[m-1][n-1];
+    }
+}