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.

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.
Example:

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.

class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
for (int i = ; i < numbers.size(); i++){
if(target - numbers[i] < numbers[i]) break; //second data smaller than current data
if(binarySearch(numbers, i+, numbers.size()-, target-numbers[i])){
vector<int> ret;
ret.push_back(i+);
ret.push_back(pos+);
return ret;
} }
} bool binarySearch(vector<int>& numbers, int start, int end, int target){
if(start > end) return false; int mid = start + (end - start )/;
if(numbers[mid] > target) return binarySearch(numbers, start, mid-, target);
else if (numbers[mid] < target) return binarySearch(numbers, mid+, end, target);
else{
pos = mid;
return true;
}
} private:
int pos; //second data index
};

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