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]. 

解法 :  

顧名思義就是要你找出這個array中有兩個數字是符合你所給定的條件，當然這一題假設只有唯一解 

解法有兩種 一個是 O(N^2) 另外一種是O(N) 

第一種就類似sort的方式，我就不用再多說了。

第二種解法的邏輯是這樣

用一個hash值儲存你到目前為止所找到的值的index，在for中每當i+1，就跟回去hash中找有沒有跟

target - nums[i] 符合的值

程式碼如下

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