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.
\n\nYour solution must use only constant extra space.
\n\n\n
Example 1:
\n\n\nInput: numbers = [2,7,11,15], target = 9\nOutput: [1,2]\nExplanation: The sum of 2 and 7 is 9. Therefore, index1 = 1, index2 = 2. We return [1, 2].\n\n\n
Example 2:
\n\n\nInput: numbers = [2,3,4], target = 6\nOutput: [1,3]\nExplanation: The sum of 2 and 4 is 6. Therefore index1 = 1, index2 = 3. We return [1, 3].\n\n\n
Example 3:
\n\n\nInput: numbers = [-1,0], target = -1\nOutput: [1,2]\nExplanation: The sum of -1 and 0 is -1. Therefore index1 = 1, index2 = 2. We return [1, 2].\n\n\n
\n
Constraints:
\n\n2 <= numbers.length <= 3 * 104-1000 <= numbers[i] <= 1000numbers is sorted in non-decreasing order.-1000 <= target <= 1000给你一个下标从 1 开始的整数数组 numbers ，该数组已按 非递减顺序排列 ，请你从数组中找出满足相加之和等于目标数 target 的两个数。如果设这两个数分别是 numbers[index1] 和 numbers[index2] ，则 1 <= index1 < index2 <= numbers.length 。
以长度为 2 的整数数组 [index1, index2] 的形式返回这两个整数的下标 index1 和 index2。
你可以假设每个输入 只对应唯一的答案 ，而且你 不可以 重复使用相同的元素。
\n\n你所设计的解决方案必须只使用常量级的额外空间。
\n \n\n示例 1：
\n\n\n输入：numbers = [2,7,11,15], target = 9\n输出：[1,2]\n解释：2 与 7 之和等于目标数 9 。因此 index1 = 1, index2 = 2 。返回 [1, 2] 。\n\n
示例 2：
\n\n\n输入：numbers = [2,3,4], target = 6\n输出：[1,3]\n解释：2 与 4 之和等于目标数 6 。因此 index1 = 1, index2 = 3 。返回 [1, 3] 。\n\n
示例 3：
\n\n\n输入：numbers = [-1,0], target = -1\n输出：[1,2]\n解释：-1 与 0 之和等于目标数 -1 。因此 index1 = 1, index2 = 2 。返回 [1, 2] 。\n\n\n
\n\n
提示：
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