LeetCode Two Sum 两数之和

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

求目标整数的两个加数，并返回其下标（C++下标+1）。

题目不难，值得注意的就是上面黄字的地方。因为可以利用这个条件稍微优化一点，使得循环由n^2编程n^2/2。循环少一半。

class Solution {
public:
	vector<int> twoSum(vector<int> &numbers, int target) {
		vector<int> twoIndices;
		if(numbers.empty()) return twoIndices;
		for(int i=0; i<numbers.size(); i++)
		{
			int temp = target - numbers[i];
			twoIndices.push_back(i+1);
			for(int k=i+1; k<numbers.size(); k++)
			{
				if(temp == numbers[k]) 
				{
					twoIndices.push_back(k+1);
					return twoIndices;
				}
			}
			twoIndices.pop_back();
		}
		return twoIndices;
	}
};

本题目leetcode上已经更新，由之前的10个testcase变成15个testcase了，可见leetcode还是在不断更新的。

更新之后，上面的程序就会超时了，所以我也更新一下程序，这次使用hash表，复杂度降低为O(n)，或者说接近O(n).

vector<int> twoSum(vector<int> &numbers, int target) 
	{
		unordered_map<int, int> mp;
		vector<int> rs(2);
		
		for (int i = 0; i < numbers.size(); i++)
		{
			int left = target - numbers[i];
			if (mp.count(left)) 
			{
				rs[0] = mp[left]+1;
				rs[1] = i+1;
				return rs;
			}
			mp[numbers[i]] = i;
		}
		return rs;
	}