[leetcode] 167. Two Sum II - Input Array Is Sorted

Two Sum II - Input Array Is Sorted

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 - 1

function twoSum(numbers: number[], target: number): number[] {
  for (let i = 0; i < numbers.length; i++) {
    for (let y = i + 1; y < numbers.length; y++) {
      if (numbers[i] + numbers[y] === target) {
        return [i + 1, y + 1];
      }
    }
  }

  return [0, 0];
}

문제풀이할때 최대한 O(n^2)의 시간복잡도를 피하기위해 중첩반복문을 피하고 싶었는데, 일단 생각나는 구현방법이 없어 부득이하게 사용하였다.