Two Sum II - Input array is sorted
Description
Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.
The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2.
Note:
Your returned answers (both index1 and index2) are not zero-based.
You may assume that each input would have exactly one solution and you may not use the same element twice.
Example:
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.
Solution
class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
int left = 0, right = numbers.size() - 1;
while(left < right){
int sum = numbers[left] + numbers[right];
if(sum < target)
++left;
else if(sum > target)
--right;
else
break;
}
return {left + 1, right + 1};
}
};
Idea of proof:
Loop invariant: it is not possible for any number in nums[0], ..., nums[left-1], nums[right+1], ..., nums[n-1]
to be one of the two numbers that sum up to target
.
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