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 = 0; i < numbers.size(); i++){ if(target - numbers[i] < numbers[i]) break; //second data smaller than current data if(binarySearch(numbers, i+1, numbers.size()-1, target-numbers[i])){ vector<int> ret; ret.push_back(i+1); ret.push_back(pos+1); return ret; } } } bool binarySearch(vector<int>& numbers, int start, int end, int target){ if(start > end) return false; int mid = start + (end - start )/2; if(numbers[mid] > target) return binarySearch(numbers, start, mid-1, target); else if (numbers[mid] < target) return binarySearch(numbers, mid+1, end, target); else{ pos = mid; return true; } } private: int pos; //second data index };