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.
思路：
two pointers法：i，j 分别指向sorted array的头和尾，如果numbers[i] + numbers[j] < target, 左指针后移，否则有指针左移。当i = j时，则没有找到和为target对应的index。则返回空。
注意：
j = numbers.length - 1; 不要再忘了-1 才是数组最后一位的index!!
代码：
public int[] twoSum(int[] numbers, int target) {
int i = 0, j = numbers.length - 1;
while (i != j) {
int sum = numbers[i] + numbers[j];
if (sum == target) {
return new int[]{i+1, j+1};
}
else if (sum > target) {
j--;
}
else {
i++;
}
}
return new int[]{};
}