문제
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:
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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:
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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:
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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 * 1041000 <= numbers[i] <= 1000numbersis sorted in non-decreasing order.1000 <= target <= 1000- The tests are generated such that there is exactly one solution.
[풀이]
1. Two Pointer 사용
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class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
left = 0
right = len(nums) - 1
while left < right:
if nums[left]+nums[right] == target:
return [left+1, right+1]
elif nums[left]+nums[right] > target:
right -=1
else:
left +=1
return []
📌주목
- Two Pointer: 리스트의 양쪽 끝에 포인터를 두고 하나씩 보는 흐름
- 단, 리스트가 정렬되어 있다는 가정 하에 문제 풀이 가능 !