# 167.two-sum-ii-input-array-is-sorted

## Statement

``````输入：numbers = [2,7,11,15], target = 9

``````输入：numbers = [2,3,4], target = 6

``````输入：numbers = [-1,0], target = -1

``````

• `2 <= numbers.length <= 3 * 104`
• `-1000 <= numbers[i] <= 1000`
• `numbers`非递减顺序 排列
• `-1000 <= target <= 1000`
• 仅存在一个有效答案

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.

## Solution

``````from typing import List

class Solution:
def twoSum(self, a: List[int], target: int) -> List[int]:
n = len(a)
k = n - 1
for i in range(n):
while k > 0 and a[i] + a[k] > target:
k -= 1
if a[i] + a[k] == target:
if i > k:
i, k = k, i
return [i + 1, k + 1]
``````