# Two Sum II

## Problem

Given a 1-indexed array of integers numbers that is already sorted in non-decreasing order, find two numbers such that they add up to a specific target number. Let these two numbers be numbers[index1] and numbers[index2] where 1 <= index1 < index2 <= numbers.length.

Return the indices of the two numbers, index1 and index2, added by one as an integer array [index1, index2] of length 2.

The tests are generated such that there is exactly one solution. You may not use the same element twice.

Your solution must use only constant extra space.

Example 1:

```
Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore, index1 = 1, index2 = 2. We return [1, 2].
```

Example 2:

```
Input: numbers = [2,3,4], target = 6
Output: [1,3]
Explanation: The sum of 2 and 4 is 6. Therefore index1 = 1, index2 = 3. We return [1, 3].
```

Example 3:

```
Input: numbers = [-1,0], target = -1
Output: [1,2]
Explanation: The sum of -1 and 0 is -1. Therefore index1 = 1, index2 = 2. We return [1, 2].
```

Constraints:

- 2 <= numbers.length <= 3 * 104
- -1000 <= numbers[i] <= 1000
- numbers is sorted in non-decreasing order.
- -1000 <= target <= 1000
- The tests are generated such that there is exactly one solution.

## Solution

```
class Solution {
public int[] twoSum(int[] numbers, int target) {
var l = 0;
var r = numbers.length - 1;
while (true) {
var sum = numbers[l] + numbers[r];
if (sum == target) {
return new int[]{l + 1, r + 1};
}
if (sum < target) {
l += 1;
} else {
r -= 1;
}
}
}
}
```