1. 题目

Given a 1-indexed array of integers numbers that is already sorted in 
non-decreasing order, find two numbers such that they add up to a specific target 
number. Let these two numbers be numbers[index1] and numbers[index2] where 1 <=
index1 < index2 <= numbers.length.
Return the indices of the two numbers, index1 and index2, added by one as an 
integer array [index1, index2] of length 2.
The tests are generated such that there is exactly one solution. You may not use 
the same element twice.
Your solution must use only constant extra space.

Example 1:

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. We return [1, 2].

Example 2:

Input: numbers = [2,3,4], target = 6
Output: [1,3]
Explanation: The sum of 2 and 4 is 6. Therefore index1 = 1, index2 = 3. We return [1, 3].

Example 3:

Input: numbers = [-1,0], target = -1
Output: [1,2]
Explanation: The sum of -1 and 0 is -1. Therefore index1 = 1, index2 = 2. We return [1, 2].

Constraints:

• 2 <= numbers.length <= 3 * 104
• -1000 <= numbers[i] <= 1000
• numbers is sorted in non-decreasing order.
• -1000 <= target <= 1000
• The tests are generated such that there is exactly one solution.

2. 解答

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

   双指针解法，可以理解为一个“元解法"。它基于排序的数组进行。利用大、小关系做一些确定性结论，从而缩小范围。
  注意“Given a 1-indexed array”，因此返回结果要在0-indexed的基础上+1。

3. 总结

  部分解法或思想需要以常识或者直觉的方式记忆在脑中，我这边称之为"元解法"。就好比数学中的乘法口诀表的存在。
  对应题干上异常的说明，要划出来。在方案写完之后，对应要点，是否都已在代码的考虑之内。