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.

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:
    vector<int> twoSum(vector<int>& numbers, int target) {
        int le = 0;
        int ri = numbers.size()-1;
        vector<int> result(0);
        while(true)
        {
            if(numbers[le] + numbers[ri] > target)
            {
                ri--;
            }else if(numbers[le] + numbers[ri] < target){
                le++;
            }else if(numbers[le] + numbers[ri] == target)
            {
                result.push_back(le+1);
                result.push_back(ri+1);
                break;
            }
        }
        return result;
    }
};