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.

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]