diff --git a/contains-duplicate/haung921209.md b/contains-duplicate/haung921209.md
new file mode 100644
index 000000000..e0b1ea984
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+
+
+# cpp stl
+```cpp
+class Solution {
+public:
+    bool containsDuplicate(vector<int>& nums) {
+        set<int> numSet(nums.begin(), nums.end());
+        return numSet.size() != nums.size();
+    }
+};
+```
+
+- set으로 단순비교. 편리하나, 정렬에 비해 시간이 오래 걸림.
+- unordered_set을 쓰면 set 방식에서 조금 더 빠를 수는 있음
+
+```cpp
+class Solution {
+public:
+    bool containsDuplicate(vector<int>& nums) {
+        sort(nums.begin(), nums.end());
+        for(int i=0;i<nums.size()-1;i++){
+            if(nums[i]==nums[i+1]){
+                return true;
+            }
+        }
+        return false;
+    }
+};
+```
+
+- 둘 다 O(n logn)의 시간 복잡도이나, 자료 특성 상 정렬이 더 빠름
+
+
+
+
+
diff --git a/house-robber/haung921209.md b/house-robber/haung921209.md
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+++ b/house-robber/haung921209.md
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+
+```cpp
+class Solution {
+public:
+    int rob(vector<int>& nums) {
+        if(nums.size()==1)
+            return nums[0];
+
+        int res = max(nums[0], nums[1]);
+        vector<int> sav(nums.size(), 0);
+        sav[0] = nums[0];
+        sav[1] = res;
+        for(int i=2;i<nums.size();i++){
+            sav[i] = max(sav[i-1], sav[i-2]+nums[i]); 
+            res = max(res, sav[i]);
+        }
+        return res;
+    }
+};
+```
+
+- 1차원 dp
+
+
diff --git a/longest-consecutive-sequence/haung921209.md b/longest-consecutive-sequence/haung921209.md
new file mode 100644
index 000000000..027b4b71a
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+++ b/longest-consecutive-sequence/haung921209.md
@@ -0,0 +1,35 @@
+```cpp
+class Solution {
+public:
+    int longestConsecutive(vector<int>& nums) {
+        if(nums.size()<=1){
+            return nums.size();
+        }
+        priority_queue<int> pq(nums.begin(), nums.end());
+        int cnt = 1;
+        int maxCnt = 1;
+        int before = pq.top();pq.pop();
+        while(pq.size()>0){
+            int cur = pq.top(); pq.pop();
+            if(before-cur ==1){
+                cnt++;
+                maxCnt = max(maxCnt, cnt);
+            }else if(before==cur){
+                continue;    
+            }else{
+                maxCnt = max(maxCnt, cnt);
+                cnt=1;
+            }
+            before = cur;
+        }
+
+        return maxCnt;
+    }
+};
+```
+
+- 순서 유지 조건 없으므로, insert와 정렬을 동시에 할 수 있는 우선순위 큐 사용
+- 전체 정렬이 필요할 경우 정렬이 더 유리할 수 있으나, 최적화 여지가 우선순위 큐가 더 커서 사용
+  - ex) 탐색 중단 조건 등
+
+
diff --git a/top-k-frequent-elements/haung921209.md b/top-k-frequent-elements/haung921209.md
new file mode 100644
index 000000000..645d0d51a
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+++ b/top-k-frequent-elements/haung921209.md
@@ -0,0 +1,23 @@
+```cpp
+class Solution {
+public:
+    vector<int> topKFrequent(vector<int>& nums, int k) {
+        unordered_map<int, int> freqCntMp;
+        vector<vector<int>> v(nums.size()+1, vector<int>());
+        int maxSize = 0;
+        for(int i=0;i<nums.size();i++){
+            freqCntMp[nums[i]]+=1;
+            v[freqCntMp[nums[i]]].push_back(nums[i]);
+            if(maxSize<freqCntMp[nums[i]])
+                maxSize = freqCntMp[nums[i]];
+        }
+        for(int idx = maxSize;idx>0;idx--){
+            if(v[idx].size()==k){
+                return v[idx];
+            }
+        }
+        return vector<int>();
+    }
+};
+```
+
diff --git a/two-sum/haung921209.md b/two-sum/haung921209.md
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+## 단순 순회
+
+```cpp
+class Solution {
+public:
+    vector<int> twoSum(vector<int>& nums, int target) {
+        int i=0,j=1;
+        for(i=0;i<j;i++){
+            for(j=i+1;j<nums.size();j++){
+                if(nums[i]+nums[j]==target){
+                    return {i, j};
+                }
+            }
+        }
+        return {0, 1};
+
+    }
+};
+```
+
+- O(n^2)
+
+## stl 사용
+
+```cpp
+class Solution {
+public:
+    vector<int> twoSum(vector<int>& nums, int target) {
+        unordered_map<int, int> valIdxMap;
+        
+        for(int idx=0;idx<nums.size();idx++){
+            if(valIdxMap.find(target-nums[idx]) != valIdxMap.end()){
+                return {idx, valIdxMap[target-nums[idx]]};
+            }
+            valIdxMap[nums[idx]] = idx;
+
+        }
+        return {0, 1};
+
+    }
+};
+```
+
+- O(nlogn)
+
+