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.

二分搜索，O(n logn）

class Solution {
    public int[] twoSum(int[] numbers, int target) {
        if(numbers == null || numbers.length < 2)
            return new int[0];

        int[] res = new int[2];
        for(int i = 0; i < numbers.length; i++) {
            int index = search(numbers, target - numbers[i]);
            if(index != -1 && index != i) {
                res[0] = index > i ? i + 1 : index + 1;
                res[1] = index > i ? index + 1 : i + 1;
                return res;
            }
        }
        return res;
    }

    public int search(int[] numbers, int target) {
        if(target < numbers[0] || target > numbers[numbers.length - 1])
            return -1;

        int left = 0;
        int right = numbers.length - 1;

        while(left < right) {
            int mid = left + (right - left) / 2;
            if(numbers[mid] == target)
                return mid;

            if(numbers[mid] > target) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }

        return numbers[left] == target ? left : -1;
    }
}

Two Pointer, O(n)，这种方法更合理一些

class Solution {
    public int[] twoSum(int[] numbers, int target) {
        if(numbers == null || numbers.length < 2)
            return new int[0];

        int[] res = new int[2];
        int left = 0;
        int right = numbers.length - 1;
        while(numbers[left] + numbers[right] != target) {
            if(numbers[left] + numbers[right] > target)
                right--;
            else
                left++;
        }

        res[0] = left + 1;
        res[1] = right + 1;

        return res;
    }
}