## Two Sum

### 描述

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: `numbers={2, 7, 11, 15}, target=9`

Output: `index1=1, index2=2`

### 分析

### 代码

{% if book.java %}
```java
// Two Sum
// 方法2：hash。用一个哈希表，存储每个数对应的下标
// Time Complexity: O(n)，Space Complexity: O(n)
public class Solution {
public int[] twoSum(int[] nums, int target) {
final HashMap myMap = new HashMap();
int[] result = new int[2];
for (int i = 0; i < nums.length; i++) {
myMap.put(nums[i],i);
}
for (int i = 0; i < nums.length; i++) {
final Integer v = myMap.get(target-nums[i]);
if (v != null && v > i) {
return new int[]{i+1, v+1};
}
}
return null;
}
};
```
{% endif %}

{% if book.cpp %}
// Two Sum
// 方法2：hash。用一个哈希表，存储每个数对应的下标
// Time Complexity: O(n)，Space Complexity: O(n)
class Solution {
public:
vector twoSum(vector &nums, int target) {
unordered_map my_map;
vector result;
for (int i = 0; i < nums.size(); i++) {
my_map[nums[i]] = i;
}
for (int i = 0; i < nums.size(); i++) {
auto iter = my_map.find(target-nums[i]);
if (iter != my_map.end() && iter->second > i) {
result.push_back(i + 1);
result.push_back(iter->second + 1);
break;
}
}
return result;
}
};
```cpp
```
{% endif %}

### 相关题目

* [3Sum](3sum.md)
* [3Sum Closest](3sum-closest.md)
* [4Sum](4sum.md)