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.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

/*
    利用hash表保存nums中的值和其下标 用target和nums中的值做差 在hash表中查找此差值是否存在
*/
class Solution
{
public:
    vector<int> twoSum(vector<int>& nums, int target)
    {
        unordered_map<int, int> hash_map;

        for (size_t i = 0; i < nums.size(); ++i)
        {
            hash_map[nums[i]] = i; // 把value和key存到hash表中  value作为键 下标作为值
        }

        vector<int> result;

        for (size_t i = 0; i < nums.size(); ++i)
        {
            const int diff = target - nums[i]; // 差值

            if (hash_map.find(diff) != hash_map.end() && hash_map[diff] > i)  // 查找diff 并且注意下标大小
            {
                result.push_back(i); // 先push_back小的值
                result.push_back(hash_map[diff]);

                break;
            }
        }

        return result;
    }
};

 

 

// 暴力 会超时 O(n^2)
class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        if (nums.empty())
        {
            return vector<int>();
        }

        vector<int> tmp;

        for (size_t i = 0; i < nums.size(); ++i)
        {
            for (size_t j = i + 1; j < nums.size(); ++j)
            {
                if (nums[i] + nums[j] == target)
                {
                    tmp.push_back(i);
                    tmp.push_back(j);

                    break;
                }
            }
        }
        return tmp;
    }
};