LeetCode | 167. Two Sum II - Input array is sorted

题目：

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.

解题思路：

这道题的目标是找到升序数组中的两个元素索引，两个元素的和等于目标值target。将两个变量变为一个定量一个变量。索引一idx1从最左边开始扫描，设定sum的值为numbers[idx1]，索引二idx2从数组的末尾开始扫描，如果numbers[idx1] + numbers[idx2]小于target，则说明以idx1为开始的所有组合都小于target，更新idx1的坐标，重新继续扫描。

题目的另一个坑是存在重复元素，需要在每次进行idx2遍历前，确定numbers[idx1]没有出现过。

代码：

public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        vector<int> res(2, 0);
        bool flag = true;
        for(int i = 0; i < numbers.size() && flag; i++)
        {
            if(i == 0 || (i > 0 && numbers[i] != numbers[i-1]))
            {
                int sum = numbers[i];
                for(int j = numbers.size() - 1; j > i; j--)
                {
                    if(sum + numbers[j] < target)
                        break;
                    if(sum + numbers[j] == target)
                    {
                        res[0] = i + 1;
                        res[1] = j + 1;
                        flag = false;
                        break;
                    }
                }
            }
        }
        return res;
    }

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