167. Two Sum II - Input array is sorted

1. Description

Given an array of integers that is already sorted in ascending order, 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.

2. Note

• Your returned answers (both index1 and index2) are not zero-based.
• You may assume that each input would have exactly one solution and you may not use the same element twice.

3. Example

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.

Example 2:
Input: numbers = [2,3,4], target = 6
Output: [1,3]

Example 3:
Input: numbers = [-1,0], target = -1
Output: [1,2]

4. Constraints

• 2 <= nums.length <= $3 * 10^4$
• -1000 <= nums[i] <= 1000
• nums is sorted in increasing order.
• -1000 <= target <= 1000

5. Solutions

My Accepted Solution

n = i_numbers.size()
Time complexity: O(n)
Space complexity: O(1)

class Solution {
public:
    vector<int> twoSum(const vector<int> &numbers, int target) {
        vector<int> indexes;
        for (int left = 1, right = numbers.size(); left < right;) {
            int sum = numbers[left - 1] + numbers[right - 1];
            if (sum < target) {
                ++left;
            } else if (sum == target) {
                indexes = {left, right};
                break;
            } else {
                --right;
            }
        }

        return indexes;
    }
};