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# 167. Two Sum II - Input Array Is Sorted (E)

https\://leetcode.com/problems/two-sum-ii-input-array-is-sorted/

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.

&#x20;

**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].
```

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**Constraints:**

* `2 <= numbers.length <= 3 * 104`
* `-1000 <= numbers[i] <= 1000`
* `numbers` is sorted in **non-decreasing order**.
* `-1000 <= target <= 1000`
* The tests are generated such that there is **exactly one solution**.

### Solution:

只要数组有序，就应该想到双指针技巧。这道题的解法有点类似二分查找，通过调节 `left` 和 `right` 可以调整 `sum` 的大小：

```java
int[] twoSum(int[] nums, int target) {
    int left = 0, right = nums.length - 1;
    while (left < right) {
        int sum = nums[left] + nums[right];
        if (sum == target) {
            // 题目要求的索引是从 1 开始的
            return new int[]{left + 1, right + 1};
        } else if (sum < target) {
            left++; // 让 sum 大一点
        } else if (sum > target) {
            right--; // 让 sum 小一点
        }
    }
    return new int[]{-1, -1};
}
```