Two Sum II - Input array is sorted

题目来源
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. 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
思路就是利用哈希表，记录遍历到的，查询target减去当前的map是否存在，直接看代码，O(n)的复杂度，感觉没有用到sorted这个一特性，应该还有更好的方法。

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        int n = numbers.size();
        vector<int> res;
        unordered_map<int, int> map;
        for (int i=0; i<n; i++) {
            if (map[target - numbers[i]]) {
                res.push_back(map[target - numbers[i]]);
                res.push_back(i+1);
                return res;
            }
            map[numbers[i]] = i + 1;
        }
        return res;
    }
};

然后发现自己傻逼了，根本没有想的那么复杂，只要两个指针，从两端向中间扫描，然后就可以了。

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        int n = numbers.size();
        int l = 0, r = n - 1;
        while (l < r) {
            if (numbers[l] + numbers[r] < target)
                l++;
            else if (numbers[l] + numbers[r] > target)
                r--;
            else
                return vector<int>({l+1, r+1});
        }
        return vector<int>();
    }
};