LeetCode #1 Two Sum

最新推荐文章于 2025-08-16 00:47:01 发布

keybord_dancer

最新推荐文章于 2025-08-16 00:47:01 发布

阅读量374

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode 文章标签： leetcode c++

本文链接：https://blog.youkuaiyun.com/keybord_dancer/article/details/47377123

leetcode 专栏收录该内容

7 篇文章

订阅专栏

本文介绍了一种使用哈希表解决两数之和问题的方法。给定一个整数数组及目标值，通过构建哈希表快速找到数组中两个数的下标，使这两个数相加等于目标值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目原文

Given an array of integers, 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

题目的意思非常简单，给定一个数组和一个数（目标值），问数组中是否有两个数的和等于目标值，如果有，给出两个数的下标。题目中已经限制了每个输入有些仅有一个答案。
这就很简单了，用哈希表就可以了。

提交代码

vector<int> twoSum(vector<int> &numbers, int target) {
        int max = INT_MIN;
        int min = INT_MAX;

        for (int i=0;i<numbers.size();i++)
        {
            if(numbers[i] >= max)
                max = numbers[i];
            if(numbers[i] <= min)
                min = numbers[i];
        }

        vector<short> hashTable(max - min + 1,0);

        for (int i=0;i<numbers.size();i++)
        {
            hashTable[numbers[i]-min] += 1;
        }

        for (int i=0;i<numbers.size();i++)
        {
            int leftTarget = target - numbers[i];
            if(leftTarget<=max && leftTarget>=min)
            {
                if ((hashTable[leftTarget-min]>0 && leftTarget != numbers[i])||
                    (leftTarget == numbers[i] && hashTable[leftTarget-min] > 1))
                {
                    vector<int> out;
                    out.push_back(i+1);
                    for (int j=0;j<numbers.size();j++)
                    {
                        if(i != j && numbers[j] == target - numbers[i])
                        {
                            out.push_back(j+1);
                        }
                    }
                    if(out[0] > out[1])
                    {
                        swap(out[0],out[1]);
                    }
                    return out;
                }   
            }
        }
        vector<int> out;
        return out;
    }