[Leetcode] 1. Two Sum

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Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

You can return the answer in any order.

Example 1:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].

Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]

Example 3:

Input: nums = [3,3], target = 6
Output: [0,1]

Constraints:

• 2 <= nums.length <= 104
• -109 <= nums[i] <= 109
• -109 <= target <= 109
• Only one valid answer exists.

Follow-up: Can you come up with an algorithm that is less than O(n2) time complexity?

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Solution 1. Brute-force

C++ 를 몰라서 vector<int>는 { , } 로 반환해야하는 것도 찾아봐야했음..

단순하게 nums 를 두번 돌면서 더한 값이 target일 때 return 시키는 방법 O(N^2)

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> ret;
        for(int i=0;i<nums.size();i++){
            for(int j=i+1;j<nums.size();j++){
                if((nums[i]+nums[j])==target){
                    return {i, j};
                }
            }
        }
        return {-1, -1};
    }
};

21년전에 python으로 풀었었는데 똑같은 brute-force의 경우 python은 3832ms, c++은 4ms,,,

이래서 c++이 성능이 좋다고 하나보다.

Solution 2. Hash Table

unordered_map을 써서 key, value 를 사용하는 방법.  

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> seen;
        for(int i=0;i<nums.size();i++){
            int diff = target-nums[i];
            if (seen.count(diff)){
                return {i, seen[diff]};
            }
            else{
                seen[nums[i]]=i;
            }
        }
        return {-1, -1};
    }
};

python으로 설명한 영상 ) https://youtu.be/luicuNOBTAI?si=TGj5FGMr1gsb_GTU