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. Please note that 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.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

利用二分法遍歷數組，如果兩數之和等於target，則把索引放入結果數組中，如果兩數之和小於target，則讓左索引+1，如果兩數之和大於target，則讓右索引-1。

class Solution {
public:
    vector<int
 
> twoSum(vector<int>& numbers, int target) {
        vector<int> res;
        int left = 0, right = numbers.size() - 1;
        while (left < right) {
            if (numbers[left] + numbers[right] == target) {
                res.push_back(left + 1);
                res.push_back(right 
 
+ 1);
            }
            if (numbers[left] + numbers[right] < target) 
                left++;
            else 
                right--;
        }
        return res;
    }
};
// 3 ms

[LeetCode] Two Sum II - Input array is sorted