167. Two Sum II - Input Array Is

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.

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.

Two Sum中的办法也可以处理，不过需要注意数组编号和大小顺序。
数组序号从0开始编号，但是返回编号是从1开始编号，且小的编号在前面。

此处考虑到是有序数组，可以考虑用二分查找的办法解决。

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        for k, v in enumerate(nums):
            new_target = target - v
            left = k + 1
            right = len(nums) - 1
            while left <= right:
                mid = left + (right-left)//2
                if nums[mid] == new_target:
                    return [k+1, mid+1] 
                elif nums[mid] < new_target:
                    left = mid + 1
                else:
                    right = mid - 1