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# 1. Two Sum

## 题目

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]
```

## 题目大意

在数组中找到 2 个数之和等于给定值的数字，结果返回 2 个数字在数组中的下标。

## 解题思路

这道题最优的做法时间复杂度是 O(n)。

顺序扫描数组，对每一个元素，在 map 中找能组合给定值的另一半数字，如果找到了，直接返回 2 个数字的下标即可。如果找不到，就把这个数字存入 map 中，等待扫到“另一半”数字的时候，再取出来返回结果。

## 代码

```cpp
class Solution {
 public:
  vector<int> twoSum(const vector<int>& nums, int target) {
    std::unordered_map<int, int> dict;
    for (int i = 0; i < nums.size(); i++) {
      int rest = target - nums[i];
      if (dict.find(rest) == dict.end()) {
        dict[nums[i]] = i;
      } else {
        return std::vector<int>{dict[rest], i};
      }
    }
    return std::vector<int>{-1, -1};
  }
};
```