前言

大家好，我是毛毛。ヾ(´∀ ˋ)ﾉ
那就開始今天的解題吧~

Question

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 * 10^4
• -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.

Think

找出陣列中，總和為target的兩個值的index。

首先~從兩個index從陣列頭跟尾逼近

1. 如果兩個index的值相加大於target就代表總和應該要減少，所以右邊的index應該要往左邊移動(減少)
2. 相反地，如果兩個index的值相加小於target就代表總和應該要增加，所以左邊的index應該要往右邊移動(增加)
3. 一直到找到兩個總和為target的值，因為題目說一定會有唯一一組解~

Code

class Solution:
    def twoSum(self, numbers: List[int], target: int) -> List[int]:
        ans = []
        
        index1 = 0
        index2 = len(numbers)-1
        
        while index1 < index2:
            total = numbers[index1] + numbers[index2]
            
            if total == target:
                return [index1+1, index2+1]
            
            elif total < target:
                index1 += 1
            
            elif total > target:
                index2 -= 1

Submission

今天就到這邊啦~
大家明天見