[LeetCode]Two Sum

最新推荐文章于 2024-03-02 21:28:00 发布

honze

最新推荐文章于 2024-03-02 21:28:00 发布

阅读量119

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/honze/article/details/38679145

LeetCode 专栏收录该内容

65 篇文章

订阅专栏

本文介绍了一个算法，用于在给定数组中找到两个数，它们相加等于特定的目标数。通过使用哈希表，我们可以在O(n)时间内解决此问题，并确保返回的索引满足题目要求。

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

这里需要注意返回的是索引，所以需要记录一下。如果有两个元素的值一样大，那么还需要注意使用哈希表存储的时候要把第一次查询到的删掉。

例如：5 2 3 2 ，target = 4时，那么返回的索引应该是 2 4

class Solution {
public:
    vector<int> twoSum(vector<int> &numbers, int target) {
        vector<int> result;
        unordered_multimap<int,int> nimap;
        for (int i = 0; i < numbers.size(); i++)
            nimap.insert(make_pair(numbers[i], i+1));
        sort(numbers.begin(), numbers.end());
        size_t i = 0, j = numbers.size() - 1;
        while(i < j)
        {
            int a = numbers[i], b = numbers[j];
            if (a + b == target)
            {
                auto ite1 = nimap.find(a);
                nimap.erase(ite1);
                auto ite2 = nimap.find(b);
                result.push_back(ite1->second);
                result.push_back(ite2->second);
                if (result[0] > result[1])
                    swap(result[0], result[1]);
                return result;
            }
            else if (a + b < target)
                ++i;
            else
                --j;
        }
        return result;
    }
};