[LeetCode] Two Sum II

Problem Description:

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.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

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The idea is very simple: set two pointers, one starts from l = 0 and the other starts from r = numbers.size() - 1. If the numbers pointed by them sum to target, we are done. Otherwise, if the their sum is larger than target, we know the number pointed by r is larger and so decrease r by 1. The case is similar if their sum is smaller than target (increase l by 1).

So we could write down the following code.

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