Leetcode之Two Sum

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

#include <iostream>
#include <vector>
#include <map>


class Solution
{
public:
	std::vector<int> twosum(std::vector<int>& nums, int target)
	{
		std::vector<int> results;
		std::map<int,int> mp;
		int len = nums.size();
		for(int i = 0; i < len; ++i)
		{
			if(!mp.count(nums[i]))
			{
				mp.insert(std::pair<int,int>(nums[i],i));
			}
			if(mp.count(target - nums[i]))
			{
				int n = mp[target - nums[i]];
				if(n < i)
				{
					results.push_back(n+1);
					results.push_back(i+1);
					return results;
				}
			}

		}
		return results;
		

	}
};

int main()
{
	int num[5] = {2,7,2,11,18};
	int target = 9;
	std::vector<int> nums(num,num+5);
	Solution sln;
	std::vector<int> results;

	results = sln.twosum(nums,target);
    int len = results.size();
	for(int i = 0; i < len; ++i)
	{
		std::cout<<results[i]<<std::endl;
	}
	return true;
	

对于Two Sum问题，首先想到的方法就是两层循环，但是提交代码后，发现超时，时间复杂度（O（n2））太高，所以必须降低时间复杂度。

我们可以利用STL的map容器来存储每个元素的索引，这样整个过程只需遍历一次就ok，而map容器的底层实现使用RB树，所以查找的时间复杂度为O（logn），这样修改代码之后，然后提交代码后，accepted，具体实现详见代码