167.TwoSumII-Inputarrayissorted.md

167. Two Sum II - Input array is sorted

• 167. Two Sum II - Input array is sorted [Array] [Two Pointers] [Binary Search] [Easy]

Tags

• [Array] [Two Pointers] [Binary Search] [Easy]

Problem

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.

Solution

class Solution {
public:
    vector<int> twoSum(vector<int> &numbers, int target) {
        vector<int> result;
        if (numbers.size() < 2) {
            return result;
        }
        
        int left = 0, right = numbers.size() - 1;
        while (left < right) {
            int sum = numbers[left] + numbers[right];
            if (sum == target) {
                result.push_back(left + 1);
                result.push_back(right + 1);
                return result;
            }
            if (sum < target){
                left++;
            } else {
                right--;
            }
        }
        
        return result;
    }
};

Time Complexity

• O(n)

Space Complexity

• O(1)

Notes

• Use Two Pointers, 20181125.

Related Problems

• 167. Two Sum II - Input array is sorted [Array] [Two Pointers] [Binary Search] [Easy]