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37 changes: 37 additions & 0 deletions contains-duplicate/haung921209.md
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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
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unordered_set을 쓰면 O(n)으로 더 빠를 수 있으나 장단점을 고려해 정렬로 풀은 점 배워갑니다. 비교해보는 것 좋네요!


```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





24 changes: 24 additions & 0 deletions 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


35 changes: 35 additions & 0 deletions longest-consecutive-sequence/haung921209.md
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```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


23 changes: 23 additions & 0 deletions top-k-frequent-elements/haung921209.md
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```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>();
}
};
```

46 changes: 46 additions & 0 deletions 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)