Problem statement:

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.

Solution:

This problem is a change version of 1. Two Sum. Given an array which is already sorted by ascending order, we need to return the index of these two numbers(actually it should be index + 1). Two pointers is the solution.

In Two Sum, we should return original indices of two numbers whose sum is equal to the target. It also works if we sort the array by ascending order(hash table to record the indices before sort).

Time complexity is O(n).

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

167. Two Sum II - Input array is sorted