```
/*
Trick : tow sum --> use pointers(it is sorred) --> compare sum left + right with target not just one side of it
Brain refresh : do not think same as TWO SUM -- > prevent your mind from thinking in another way
*/
public int[] twoSum(int[] numbers, int target) {
// same number should not use twice
int left = 0; int right = numbers.length -1;
while (left < right) {
int sum = numbers[left] + numbers[right];
if (sum < target) {
left ++;
}else if (sum > target) {
right--;
}else {
return new int[]{left + 1, right +1};
}
}
return new int[2];
}
```

In this problem , I first think in TWO SUM ways --> but use binary search instead of map. However, the efficiency is O(nLogn). So I came up with solution using two pointers and sorted array property well.