LeetCode 167. Two Sum II - Input array is sorted （夹逼法）

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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。采用双指针技术，从两端向中间逼近，实现O(n)的时间复杂度。

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

输入一组整数（升序排列）和一个整数，找到数组中相加等于这个整数的两个数的序号。有且只有一组解。

思路：一开始使用两层循环，但是超时。

    vector<int> twoSum(vector<int>& numbers, int target) {
        int a,b;
        for(a=0;a<numbers.size();a++)
        {
            for(b=numbers.size()-1;b>a;b--)
            {
                if(numbers[a]+numbers[b]==target)
                    return vector<int> {a+1,b+1};
            }
        }
    }

改用夹逼法。

    vector<int> twoSum(vector<int>& numbers, int target) {
        int a=0,b=numbers.size()-1;
        while(a<b)
        {
            if(numbers[a]+numbers[b]==target)
                return vector<int> {a+1,b+1};
            else if (numbers[a]+numbers[b]<target)
                a++;
            else
                b--;
        }
    }