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] <= 1000numbersis sorted in non-decreasing order.-1000 <= target <= 1000- The tests are generated such that there is exactly one solution.
思路
双指针，分别初始化于数组的最左/右边，将左指针与右指针的值相加与目标值比较，如果目标值比较大，则左指针进一，反之右指针退一，如此循环直至相等，此时左指针和右指针的位置即为答案。
代码
python版本：
class Solution:
def twoSum(self, numbers: List[int], target: int) -> List[int]:
l, r = 0, len(numbers)-1
while True:
add = numbers[l]+numbers[r]
if add > target:
r -= 1
elif add < target:
l += 1
else:
return [l+1, r+1]
