167. Two Sum II - Input array is sorted

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本文介绍了一种在已排序数组中寻找两个数使它们的和等于特定目标值的方法。提出了两种解决方案：一种利用二分查找，另一种使用对撞指针技术。这两种方法都充分利用了输入数组已经排序的特点。

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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1：暴力解法，双层循环遍历O(n^2)
暴力枚举是思考的开始

思路2：数组的性质 Input array is sorted 有序 –> 二分查找(target-numbers[i])

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        int l = 0, r = numbers.size() - 1, position = -1, i = 0;
        for (i = 0; i < numbers.size(); i++) {
            int searchNum = target - numbers[i];
            l = i + 1;
            while (l <= r)
            {
                int mid = l + ((r - l) >> 1);
                if (searchNum == numbers[mid]) {
                    position = mid;
                    break;
                }
                else if (searchNum < numbers[mid]) {
                    r = mid - 1;
                }
                else {
                    l = mid + 1;
                }
            }

            if (position != -1) {
                if (i != position) {
                    int res[] = {i+1, position+1};
                    return vector<int>(res, res + 2);
                }
            }
        }


        throw invalid_argument("The input has no solution.");
    }
};

思路2：对撞指针
数组无序就不能采用这种思路

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        int l = 0, r = numbers.size() - 1;

        while (l < r) {
            int sum = numbers[l] + numbers[r];
            if (sum == target) {
                int res[] = {l+1, r+1};
                return vector<int>(res, res + 2);
            } else if (sum < target) {
                ++l;
            } else {
                --r;
            }
        }


        throw invalid_argument("The input has no solution.");
    }
};