# 两数之和
<p>给定一个整数数组 <code>nums</code> 和一个整数目标值 <code>target</code>，请你在该数组中找出 <strong>和为目标值</strong> 的那 <strong>两个</strong> 整数，并返回它们的数组下标。</p><p>你可以假设每种输入只会对应一个答案。但是，数组中同一个元素在答案里不能重复出现。</p><p>你可以按任意顺序返回答案。</p><p> </p><p><strong>示例 1：</strong></p><pre><strong>输入：</strong>nums = [2,7,11,15], target = 9<strong><br />输出：</strong>[0,1]<strong><br />解释：</strong>因为 nums[0] + nums[1] == 9 ，返回 [0, 1] 。</pre><p><strong>示例 2：</strong></p><pre><strong>输入：</strong>nums = [3,2,4], target = 6<strong><br />输出：</strong>[1,2]</pre><p><strong>示例 3：</strong></p><pre><strong>输入：</strong>nums = [3,3], target = 6<strong><br />输出：</strong>[0,1]</pre><p> </p><p><strong>提示：</strong></p><ul>	<li><code>2 <= nums.length <= 10<sup>3</sup></code></li>	<li><code>-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup></code></li>	<li><code>-10<sup>9</sup> <= target <= 10<sup>9</sup></code></li>	<li><strong>只会存在一个有效答案</strong></li></ul>
<p><p>给定一个整数数组 <code>nums</code> 和一个整数目标值 <code>target</code>，请你在该数组中找出 <strong>和为目标值</strong> 的那 <strong>两个</strong> 整数，并返回它们的数组下标。</p><p>你可以假设每种输入只会对应一个答案。但是，数组中同一个元素在答案里不能重复出现。</p><p>你可以按任意顺序返回答案。</p><p> </p><p><strong>示例 1：</strong></p><pre><strong>输入：</strong>nums = [2,7,11,15], target = 9<strong><br />输出：</strong>[0,1]<strong><br />解释：</strong>因为 nums[0] + nums[1] == 9 ，返回 [0, 1] 。</pre><p><strong>示例 2：</strong></p><pre><strong>输入：</strong>nums = [3,2,4], target = 6<strong><br />输出：</strong>[1,2]</pre><p><strong>示例 3：</strong></p><pre><strong>输入：</strong>nums = [3,3], target = 6<strong><br />输出：</strong>[0,1]</pre><p> </p><p><strong>提示：</strong></p><ul>	<li><code>2 <= nums.length <= 10<sup>3</sup></code></li>	<li><code>-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup></code></li>	<li><code>-10<sup>9</sup> <= target <= 10<sup>9</sup></code></li>	<li><strong>只会存在一个有效答案</strong></li></ul>
<p>以下错误的选项是？</p></p>
## aop
### before
```cpp
#include <unordered_map>
#include <vector>
#include <iostream>
#include <map>
using namespace std;
```
### after
```cpp
int main()
{
	Solution test;
	int arr[] = {3,2,4};
	vector<int> ret;
	vector<int> vec(arr, arr+4);
	ret = test.twoSum(vec, 6);
	for (auto i = ret.begin(); i != ret.end(); i++)
	{
		cout << *i << ' ';
	}
	return 0;
}
```

## 答案
```cpp
class Solution
{
public:
    vector<int> twoSum(vector<int> &nums, int target)
    {
        vector<int> vec;
        map<int, int> dic; 
        for (int i = 0; i < nums.size(); ++i)
        {
            dic[nums[i]] = i;
        }
        for (int i = 0; i < nums.size(); ++i)
        {
            if (dic.count(target - nums[i]) != 0 && dic[target - nums[i]] != i)
            {
                vec.push_back(i);
                vec.push_back(dic[target]);
                break;
            }
        }
        return vec;
    }
};
```
## 选项

### A
```cpp
class Solution
{
public:
    vector<int> twoSum(vector<int> &nums, int target)
    {
        vector<int> a;
        map<int, int> map;
        for (int i = 0; i < nums.size(); i++)
        {
            map[nums[i]] = i;
        }
        for (int j = 0; j < nums.size(); j++)
        {
            if (map.count(target - nums[j]) == 1 && map[target - nums[j]] != j)
            {
                a.push_back(j);
                a.push_back(map[target - nums[j]]);
                return a;
            }
        }
        return a;
    }
};
```

### B
```cpp
class Solution
{
public:
    vector<int> twoSum(vector<int> &nums, int target)
    {
        vector<int> a;
        map<int, int> map;
        for (int i = 0; i < nums.size(); i++)
        {
            if (map.find(target - nums[i]) != map.end())
            {
                a.push_back(map[target - nums[i]]);
                a.push_back(i);
                return a;
            }
            else
            {
                map[nums[i]] = i;
            }
        }
        return a;
    }
};
```

### C
```cpp
class Solution
{
public:
    vector<int> twoSum(vector<int> &nums, int target)
    {
        vector<int> a;
        for (int i = 0; i < nums.size(); i++)
        {
            for (int j = i + 1; j < nums.size(); j++)
            {
                if (nums[i] + nums[j] == target)
                {
                    a.push_back(i);
                    a.push_back(j);
                    return a;
                }
            }
        }
    }
};
```