力扣-167-有序数组的TwoSum

题目：
167. Two Sum II - Input array is sorted(easy)
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.

1 2 3 4

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.

题目大意：
给定一个已按照升序排列 的有序数组，找到两个数使得它们相加之和等于目标数。
函数应该返回这两个下标值 index1 和 index2，其中 index1 必须小于 index2。

说明:
返回的下标值（index1 和 index2）不是从零开始的。
你可以假设每个输入只对应唯一的答案，而且你不可以重复使用相同的元素。

解题思路：
在有序数组中使用双指针，第一个指针指向值较小的数，第二个指针指向值较大的数。
第一个指针从前向后遍历，第二个指针从后向前遍历。

代码：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16

Java:
public int[] twoSum(int[] numbers,int target){
    if(numbers == null) return null;
    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){
            i ++;
        }else{
            j --;
        }   
    }
    return null;
}