the description:
Given an array of integers <code>nums</code> and an integer <code>target</code>, return indices of the two numbers such that they add up to the 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.
the first step:
understanding the issue:
the second step:
code writing
<pre><code class="lang-cpp">class Solution {
 public int[] twoSum(int[] nums, int target) {
 Hashtable&lt;Integer, Integer&gt; hashTable = new Hashtable&lt;Integer, Integer&gt;();
 int i = 0;
 while ((i &lt; nums.length) &amp;&amp; (hashTable.get(nums[i]) == null)) { 
 hashTable.put(target - nums[i], i);
 i++;
 }
 if (i &lt; nums.length) {
 return new int[]{hashTable.get(nums[i]),i};
 } 
 return null; 
 }
};
</code></pre>

the description:

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

the first step:

understanding the issue:

the second step:

code writing

```cpp
class Solution {
 public int[] twoSum(int[] nums, int target) {
 Hashtable<Integer, Integer> hashTable = new Hashtable<Integer, Integer>();
 int i = 0;
 while ((i < nums.length) && (hashTable.get(nums[i]) == null)) { 
 hashTable.put(target - nums[i], i);
 i++;
 }
 if (i < nums.length) {
 return new int[]{hashTable.get(nums[i]),i};
 } 
 return null; 
 }
};
```

solve problems| Two Sum