Two Sum II - Input array is sorted

最新推荐文章于 2025-08-22 19:21:44 发布

原创最新推荐文章于 2025-08-22 19:21:44 发布 · 206 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#C++ #leetcode

LeetCode 专栏收录该内容

161 篇文章

订阅专栏

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，从数组两端开始向中间搜索，实现了高效的查找。文章还提供了一个C++示例代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

一、问题描述

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. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

二、思路

由于本题只有两个数，而且数组是有序的，所以根据二分查找的思路即可，如果找到等于target的两个数，直接返回下标；找不到直接返回空。

三、代码

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        int p1 = 0, p2 = numbers.size() - 1;
        while(p1 < p2){
            if(numbers[p1] + numbers[p2] > target) --p2;
            else if(numbers[p1] + numbers[p2] < target) ++p1;
            else return vector<int>({p1 + 1, p2 + 1});
        }
        return vector<int>();
    }
};