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. 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 解法一：暴力搜索 当前数字依次与后续数字求和，结果等于target就输出位置。此方法虽然可以得到结果，但是时间复杂度高，会超时。 vector<int> twoSum(vector<int>& numbers, int target) { vector<int> result(2,0); for(int i = 0; i < numbers.size(); i++) for(int j = i + 1; j < numbers.size(); j++) { if(numbers[i] + numbers[j] == target) { result[0] = i + 1; result[1] = j + 1; } } return result; }解法二：二分查找 依次遍历数组中每一个数字numbers[i]，采用二分查找搜索target - numbers[i]，ac vector<int> twoSum(vector<int>& numbers, int target) { vector<int> result(2,0); int position = -1, low = 0, high = numbers.size() - 1; for(int i = 0; i < numbers.size(); i++) { low = 0; high = numbers.size() - 1; int k = target - numbers[i]; while(low <= high) { if(numbers[(low + high)/2] == k) { if(i == (low + high)/2) low = low = (low + high)/2 + 1; else { position = (low + high)/2; break; } } if(numbers[(low + high)/2] < k) low = (low + high)/2 + 1; if(numbers[(low + high)/2] > k) high = (low + high)/2 - 1; } if(position != -1) { result[0] = i + 1; result[1] = position + 1; break; } } return result; }