leetcode 167. Two Sum II - Input array is sorted-优快云博客

本文链接：https://blog.youkuaiyun.com/level_code/article/details/94911652

本文介绍了一种解决有序数组中寻找两个数使其和为目标值的问题的方法。利用双指针技巧，从数组两端开始，根据当前和与目标值的比较调整指针位置，最终找到符合条件的两个数并返回其下标。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

two sum的问题，与第1题的two sum区别是已经是sorted的数组

思路：
因为数组已经排好序，可以用双指针的方法。
引用two pointers的思想，指针指向两端，和 < target的时候右端指针向左移，否则左边指针向右移，因为题目中说有唯一解，所以当和=target的时候直接返回下标（以1为起点）

//0ms
    public int[] twoSum(int[] numbers, int target) {
        if (numbers == null || numbers.length == 0) {
            return new int[]{};
        }
        
        int[] res = new int[2];
        int i = 0;
        int j = numbers.length - 1;
        
        while (i < j) {
            if (numbers[i] + numbers[j] == target) {
                res[0] = i + 1;
                res[1] = j + 1;
                return res;
            } else if (numbers[i] + numbers[j] < target) {
                i ++;
            } else {
                j --;
            }
        }
        
        return res;
    }