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.

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1）

由于数组是已经排序好的，所以用二分查找，时间复杂度为O(nlogn)。就是先遍历数组，然后查找这个数的右边的是否有一个数，这个数与它相加得到目标数。

C++代码：

class Solution {

public:

    vector<int> twoSum(vector<int>& numbers, int target) {

        for(int i = ;i < numbers.size(); i++){

            int t = target - numbers[i],left = i + ,right = numbers.size() - ;

            while(left <= right){

                int mid = left + (right - left)/;

                if(numbers[mid] == t) return {i+,mid+};

                else if(numbers[mid] < t) left = mid + ;

                else right = mid - ;

            }

        }

        return {};

    }

};

2）

不过二分查找的时间复杂度比较大，可以用双指针，时间复杂度为线性。空间复杂度为O(1)。

C++代码：

class Solution {

public:

    vector<int> twoSum(vector<int>& numbers, int target) {

        int l = ,r = numbers.size() - ;

        while(l < r){

            if(numbers[l] + numbers[r] == target) return {l+,r+};

            else if(numbers[l] + numbers[r] > target) r--;

            else l++;

        }

        return {};

    }

};