Two Sum II
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 1 as an integer array 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.
Constraints:
- 2 <= numbers.length <= 1000
- -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.
Examples:
Input: [2,7,11,15] and target = 9
Output: [1,2]
Explanation: Because numbers[0] + numbers[1] == 2 + 7 == 9, we return [1, 2].
Solutions
Two Pointers
We use two pointers, one at the start and one at the end of the array. We calculate the sum of the numbers at these two pointers. If the sum is equal to the target, we return the indices of these two numbers. If the sum is less than the target, we move the left pointer to the right. If the sum is greater than the target, we move the right pointer to the left.
def twoSum(numbers, target):
left, right = 0, len(numbers) - 1; while left < right:
sum_ = numbers[left] + numbers[right]; if sum_ == target:
return [left + 1, right + 1]; elif sum_ < target:
left += 1; else:
right -= 1Follow-up:
What if the array is not sorted?
