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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
def twoSum1(self, numbers, target):
l, r = 0, len(numbers)-1
while l < r:
s = numbers[l] + numbers[r]
if s == target:
return [l+1, r+1]
elif s < target:
l += 1
else:
r -= 1
public int[] twoSum(int[] num, int target) {
int[] indice = new int[2];
if (num == null || num.length < 2) return indice;
int left = 0, right = num.length - 1;
while (left < right) {
int v = num[left] + num[right];
if (v == target) {
indice[0] = left + 1;
indice[1] = right + 1;
break;
} else if (v > target) {
right --;
} else {
left ++;
}
}
return indice;
}
vector<int> twoSum(vector<int>& numbers, int target) {
int l = 0;
int r = numbers.size() -1;
while(l < r){
if(numbers[l] + numbers[r] == target){
return {l+1,r+1};;
}
else if(numbers[l] + numbers[r] > target){
r--;
}
else{
l++;
}
}
return {};
}
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