LeetCode: 2 Sum (Input array is sorted)

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本文介绍了一种在已排序的整数数组中寻找两个数使其和等于特定目标值的有效算法。采用首尾指针的方法，时间复杂度为O(n)，确保了高效性和准确性。

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

思路：首尾两个指针遍历即可,时间复杂度 O(n) 。

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