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.
풀이
def sum(lst,count,new_lst,idx):
if not lst:
return []
if count == 0:
return new_lst
for i in range(idx,len(lst)):
sum(lst, count - lst[i], new_lst + [i + 1], i + 1)
result = []
sum(numbers,target,result,0)
처음에는 재귀 함수 이용하여 new_lst에 lst[i]에 값을 추가하면서 count에 lst[i]에 값을 빼주는 방법을 생각했다.
그런데 count가 0이 되었는데도 재귀함수가 종료되지 않고 계속 재귀함수가 호출되어서
result = []
left, right = 0, len(numbers)-1
while left <= right:
cur = numbers[right] + numbers[left]
if target == cur:
result.append(left+1)
result.append(right+1)
break
elif target < cur:
right -=1
else:
left +=1
return result
투포인터를 이용해 배열이 오름차순으로 정렬되어있으므로 cur가 target보다 클 때는 cur의 크기를 줄여야하기때문에 right를 1 감소, cur가target보다 작을때는 left를 1 증가하여 cur의 총합이 증가 할 수 있도록 하였고 cur와 target의 크기가 같을때 left와 right를 1 증가시켜 result 배열에 추가
