diff --git a/container-with-most-water/Geegong.java b/container-with-most-water/Geegong.java
new file mode 100644
index 0000000000..d462873902
--- /dev/null
+++ b/container-with-most-water/Geegong.java
@@ -0,0 +1,84 @@
+public class Geegong {
+
+    /**
+     * two pointer 를 사용, 처음에는 height 배열을 정렬해서 높은 순오르 최대 면적을 구하는건가 싶었지만 그럴 필요는 없었음
+     * (어차피 모든 원소를 돌아야되기 때문에 소팅의 의미가 없음, 그리고 NLogN 시간 복잡도가 생겨서 더 좋지 않음)
+     * time complexity : O(N)
+     * space complexity : O(N)
+     *
+     * @param height
+     * @return
+     */
+    public int maxArea(int[] height) {
+
+
+        int leftIdx=0;
+        int rightIdx = height.length - 1;
+        int maxVolume = 0;
+
+        while(leftIdx < rightIdx) {
+
+            int leftHeight = height[leftIdx];
+            int rightHeight = height[rightIdx];
+            int gap = rightIdx - leftIdx;
+            int currentVolume = gap * Math.min(leftHeight, rightHeight);
+            maxVolume = Math.max(currentVolume, maxVolume);
+
+            if (leftHeight > rightHeight) {
+                rightIdx--;
+            } else {
+                leftIdx++;
+            }
+
+        }
+
+        return maxVolume;
+
+
+
+
+
+
+
+
+////////// 아래 부분은 예전 기수에서 풀었던 방법
+//        int leftIndex = 0;
+//        int rightIndex = height.length - 1;
+//
+//        int maxAmount = 0;
+//        int currentAmount = 0;
+//
+//        while(leftIndex != rightIndex && leftIndex < rightIndex) {
+//            // 면적을 먼저 구해본다.
+//            int minHeight = Math.min(height[leftIndex], height[rightIndex]);
+//            currentAmount = minHeight * (rightIndex - leftIndex);
+//
+//            maxAmount = Math.max(currentAmount, maxAmount);
+//            // 어느 포인터를 움직일지 결정
+//            /**
+//             * case 1. 단순히 전체를 모두 훑어버리면 Time limit exceeded 발생
+//             */
+////            if (leftIndex < rightIndex - 1) {
+////                rightIndex--;
+////            } else if (leftIndex == rightIndex - 1) {
+////                rightIndex = height.length - 1;
+////                leftIndex++;
+////            }
+//
+//            /**
+//             * case 2. 포인터가 돌면서 높은 height만 고려해서 포인터가 움직일 때에는 Time limit exceeded 발생 X
+//             */
+//            if (height[leftIndex] < height[rightIndex]) {
+//                leftIndex++;
+//            } else if (height[leftIndex] > height[rightIndex]) {
+//                rightIndex--;
+//            } else {
+//                rightIndex--;
+//                leftIndex++;
+//            }
+//        }
+//
+//        return maxAmount;
+    }
+}
+
diff --git a/jump-game/Geegong.java b/jump-game/Geegong.java
new file mode 100644
index 0000000000..17c81ae263
--- /dev/null
+++ b/jump-game/Geegong.java
@@ -0,0 +1,56 @@
+public class Geegong {
+
+    /**
+     * greedy algorithms 사용
+     *
+     * time complexity : O(N)
+     * space complexity : O(N)
+     * @param nums
+     * @return
+     */
+    public boolean canJump(int[] nums) {
+        // greedy algorithm
+        int farthest = 0;
+        int lastIdx = nums.length - 1;
+        for (int idx=0; idx<nums.length; idx++) {
+            // idx가 가장 멀리갈 수 있는 거리보다도 넘었다면 가망이 없는것
+            if (idx > farthest) return false;
+            int currentFarthest = idx + nums[idx];
+            farthest = Math.max(farthest, currentFarthest);
+
+            if (farthest >= lastIdx) return true;
+        }
+
+        return true;
+    }
+
+
+    // 아래 방법으로 풀면 TLE 발생
+//    public boolean canJump(int[] nums) {
+//        return recursion(0, nums);
+//    }
+//
+//    public boolean recursion(int idx, int[] nums) {
+//        int totalLength = nums.length;
+//        if (idx == totalLength - 1) {
+//            return true;
+//        }
+//
+//        if (idx >= totalLength) {
+//            return false;
+//        }
+//
+//        int currVal = nums[idx];
+//        while(currVal > 0) {
+//            boolean result = recursion(idx + currVal, nums);
+//            if (result) {
+//                return true;
+//            }
+//
+//            currVal--;
+//        }
+//
+//        return false;
+//    }
+
+}
diff --git a/merge-two-sorted-lists/Geegong.java b/merge-two-sorted-lists/Geegong.java
new file mode 100644
index 0000000000..3c40ab6b8a
--- /dev/null
+++ b/merge-two-sorted-lists/Geegong.java
@@ -0,0 +1,45 @@
+public class Geegong {
+
+    /**
+     * 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; }
+     * }
+     */
+    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
+
+        ListNode root = new ListNode(Integer.MIN_VALUE, list1);
+        ListNode curr1 = list1;
+        ListNode curr2 = list2;
+
+        while (curr1 != null && curr2 != null) {
+            int val1 = curr1.val;
+            int val2 = curr2.val;
+
+            if (val1 <= val2) {
+                ListNode temp = curr1.next;
+                curr1.next = curr2;
+                curr1 = temp;
+            } else if (val1 > val2) {
+                ListNode temp = curr2.next;
+                curr2.next = curr1;
+                curr2 = temp;
+            }
+        }
+
+        return root.next;
+    }
+
+    public static class ListNode {
+          int val;
+          ListNode next;
+          ListNode() {}
+          ListNode(int val) { this.val = val; }
+          ListNode(int val, ListNode next) { this.val = val; this.next = next; }
+    }
+
+}
diff --git a/sum-of-two-integers/Geegong.java b/sum-of-two-integers/Geegong.java
new file mode 100644
index 0000000000..1cbddca0c3
--- /dev/null
+++ b/sum-of-two-integers/Geegong.java
@@ -0,0 +1,25 @@
+public class Geegong {
+    /**
+     * XOR, AND 연산을 이용하여 계산
+     * XOR : carry 확인하여 1비트씩 왼쪽으로 이동 (<<)
+     * AND : carry 없이 더하기 결과값만 확인
+     * time complexity : O(N)
+     * space complexity : O(N)
+     * @param a
+     * @param b
+     * @return
+     */
+    public int getSum(int a, int b) {
+        int a_ = a;
+        int b_ = b;
+
+        while (a_ != 0) {
+            int temp = a_;
+            a_ = (a_ & b_) << 1;
+            b_ = (temp ^ b_);
+        }
+
+        return b_;
+    }
+
+}