1 Description #
source: https://leetcode.com/problems/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 * 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.
2 Solution #
#include <algorithm>
class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
// Space complexity: O(1)
// Time complexity: O(N)
int low = 0;
int high = numbers.size() - 1;
while(numbers[low] + numbers[high] != target){
if(numbers[low] + numbers[high] < target){
low ++;
}else{
high--;
}
}
std::vector<int> result{low + 1, high + 1};
return result;
}
};