# LeetCode 1. Two Sum

Given an array of integers `nums` and an integer `target`, return *indices of the two numbers such that they add up to `target`*.

You may assume that each input would have ***exactly***** one solution**, and you may not use the *same* element twice.

You can return the answer in any order.

**Example 1:**

```
Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Output: Because nums[0] + nums[1] == 9, we return [0, 1].
```

**Example 2:**

```
Input: nums = [3,2,4], target = 6
Output: [1,2]
```

**Example 3:**

```
Input: nums = [3,3], target = 6
Output: [0,1]
```

**Constraints:**

* `2 <= nums.length <= 103`
* `-109 <= nums[i] <= 109`
* `-109 <= target <= 109`
* **Only one valid answer exists.**

## Solution:

[English Version in Youtube](https://youtu.be/WbSOR-Qewt0)

[中文版解答Youtube Link](https://youtu.be/GKM0IByXKkk)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV1dy4y1n7AS/)

### Solution 1: Double for loop

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

Time Complexity: O(n^2)

Space Complexity: O(1)

### Solution 2: Hash Map

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

Time Complexity: O(n)

Space Complexity: O(n)


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://zhenchaogan.gitbook.io/leetcode-solution/leetcode-1-two-sum.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.