[leetcode-1-Two Sum]

Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

    vector<int> twoSum(vector<int>& nums, int target)
    {
        if (nums.size() == 0) return nums;
        vector<int>result;
        //sort(nums.begin(), nums.end());
        for (int i = 0; i < nums.size(); i++)
        {
            for (int j = i + 1; j < nums.size();j++)
            {
                if (nums[i]+nums[j] == target)
                {
                    result.push_back(i);
                    result.push_back(j);
                    return result;
                }
            }
        }
        return result;
    }

然后看到大牛的代码：只需遍历一次数组，时间复杂度为O(n).

vector<int> twoSum(vector<int> &numbers, int target)
{
    //Key is the number and value is its index in the vector.
    unordered_map<int, int> hash;
    vector<int> result;
    for (int i = 0; i < numbers.size(); i++) {
        int numberToFind = target - numbers[i];

            //if numberToFind is found in map, return them
        if (hash.find(numberToFind) != hash.end()) {
                    //+1 because indices are NOT zero based
            result.push_back(hash[numberToFind] + 1);
            result.push_back(i + 1);            
            return result;
        }

            //number was not found. Put it in the map.
        hash[numbers[i]] = i;
    }
    return result;
}