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
     
       &a, const pair
      
        &b) {    return a.first < b.first;}class Solution {public:    vector
       
         twoSum(vector
        
          &numbers, int target) {        return twoSum_2(numbers, target);    }        // solution 1: sort and select O(nlgn)    vector
         
           twoSum_1(vector
          
            &numbers, int target) { // pair
           
             vector
            
             
               > 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
              
                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
               
                 twoSum_2(vector
                
                  &numbers, int target) { // exist bug; hash-table unordered_map
                 
                   map; for (int i = 0; i < numbers.size(); ++i) map[numbers[i]] = i + 1; for (int i = 0; i < numbers.size(); ++i) { unordered_map
                  
                   ::iterator it = map.find(target - numbers[i]); if (it == map.end() || i+1 == it->second) continue; vector
                   
                     res; res.push_back(min(i+1, it->second)); res.push_back(max(i+1, it->second)); return res; } }};