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Algorithm/Leet Code

LeetCode #260 SingleNumber3. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접,연결리스트

LeetCode #260 SingleNumber3. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접,연결리스트

 

 

Runtime: 20 ms, faster than 22.92% of C++ online submissions for Single Number III.

 

Memory Usage: 10.7 MB, less than 20.69% of C++ online submissions for Single Number III.

 

 

LeetCode #260

Q.

 Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.

 

 주어진 숫자 배열에서, 딱 두개의 값만 한번씩 나오고 나머지 값들은 전부 두번씩 나온다. 그 딱 한번 나오는 두개의 값을 찾아봐라.

 

 

Example:

 

Input: [1,2,1,3,2,5]

Output: [3,5]

 

Note:

  1. The order of the result is not important. So in the above example, [5, 3] is also correct.
  2. Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

 1. 순서는 중요하지 않다. 3, 5 나 5, 3 이나 둘다 답

 2. 선형의 시간복잡도가 되도록, 상수 공간복잡도로 구현.

 

Process

// Process 
//1. Input array

//2. Pick twice elem using map

//3. Move once elem to vector

//4. Return vector

 

 

// 처리과정

//1. 배열 입력받는다

//2. 맵에서 2개인 값을 걸러낸다.

//3. 벡터로 옮긴다.

//4. 벡터 반환한다.

 

 

Code.. lemme see example code!!!

코드.. 예제코드를 보자!!!

 

 

 

class Solution {
public:
	vector<int> singleNumber(vector<int>& nums) {
		vector<int> result;

		map<int, int> buffer;
		for (int i = 0; i < nums.size(); ++i) {
			if (buffer.count(nums[i]) > 0) {
				buffer.erase(nums[i]);
			}
			else {
				buffer.insert(make_pair(nums[i], 0));
			}
		}
		map<int, int>::iterator it;
		for (it = buffer.begin(); it != buffer.end(); ++it) {
			result.push_back(it->first);
		}
        
		return result;
	}
};

 

 

 

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