Two Sum

最新推荐文章于 2025-04-09 16:12:02 发布

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本文介绍了一种解决两数之和问题的有效算法。通过使用哈希表存储数组中的每个元素及其对应下标，能够在O(n)的时间复杂度内找到两个数，使它们的和等于目标值。该算法确保了每个输入都有唯一解，并返回这两个数的非零基下标。

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

hash。用一个哈希表，存储每个数对应的下标,时间复杂度 O(n)，空间复杂度 O(n)

class Solution 
{
public:
	vector<int> twoSum(vector<int> &num, int target)
	{
		unordered_map<int> mapping;
		vector<int, int> result;
//把mapping[num[i]] = i
		for (int i = 0;i < num.size();++i)
		{
			mapping[num[i]] = i;
		}
		for (int i = 0;i < num.size();++i)
		{
			const int gap = target - num[i];
			if (mapping.find(gap) != num.end())
			{
				result.push_back(i + 1);
				result.push_back(num[i]+1);
			}
		}
		return result;
	}
};