2Sum

2Sum

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target,
where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

bool compare(const pair<int, int> &a, const pair<int, int> &b) {
    return a.first < b.first;
}

class Solution {
public:
    vector<int> twoSum(vector<int> &numbers, int target) {
        return twoSum_2(numbers, target);
    }
    
    // solution 1: sort and select O(nlgn)
    vector<int> twoSum_1(vector<int> &numbers, int target) { // pair<val,index>
        vector<pair<int, int> > nums(numbers.size());
        for (int i = 0; i < numbers.size(); ++i)
            nums[i] = make_pair(numbers[i], i+1);
        sort(nums.begin(), nums.end(), compare);
        
        int l = 0, r = nums.size() - 1;
        while (l < r) {
            int sum = nums[l].first + nums[r].first;
            if (sum == target) break;
            else if (sum < target) l++;
            else r--;
        }

        vector<int> res;
        res.push_back(min(nums[l].second, nums[r].second));
        res.push_back(max(nums[l].second, nums[r].second));
        return res;
    }
    
    // solution 2: hashtable O(n)
    vector<int> twoSum_2(vector<int> &numbers, int target) { // exist bug; hash-table
        unordered_map<int, int> map;
        for (int i = 0; i < numbers.size(); ++i)
            map[numbers[i]] = i + 1;
        for (int i = 0; i < numbers.size(); ++i) {
            unordered_map<int, int>::iterator it = map.find(target - numbers[i]);
            if (it == map.end() || i+1 == it->second) continue;
            vector<int> res;
            res.push_back(min(i+1, it->second));
            res.push_back(max(i+1, it->second));
            return res;
        }
    }
};