twoSum

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

 

第一次的代码，思路很简单：

#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

class Solution{
public:
        vector<int> twoSum(vector<int>& nums, int target){
                int i = 0;
                int j = 0;
                vector<int> result;

                for(i=0; i<nums.size()-1; i++){
                        for(j=i+1; j<nums.size(); j++){
                                if(nums[i] + nums[j] == target){
                                        result.push_back(i+1);
                                        result.push_back(j+1);
                                }
                        }
                }
                return result;
        }
};

int main(){
        vector<int> src;
        int input;
        int target;
        int i =0;

        vector<int> result;

        do{
                cin>>input;
                src.push_back(input);
                if(cin.get() == '
'){
                        break;
                }
        }while(1);

        cin>>target;

        Solution s;
        result = s.twoSum(src, target);

        for(i=0; i<result.size(); i++){
                cout<<result[i]<<endl;
        }

        return 0;
}

最后发现，由于复杂度是O(n^2)系统无法接受，需要降低算法复杂度。。。

现在看到别人用map来解决这个问题   

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

using namespace std;

class Solution {
public:
	vector<int> twoSum(vector<int> &numbers, int target) {
		vector<int> result;
		map<int, int> m;
		if (numbers.size() < 2)
			return result;
		for (int i = 0; i < numbers.size(); i++)
			m[numbers[i]] = i;

		map<int, int>::iterator it;
		for (int i = 0; i < numbers.size(); i++) {
			if ((it = m.find(target - numbers[i])) != m.end())
			{
				if (i == it->second) continue;
				result.push_back(i+1);
				result.push_back(it->second+1);
				return result;
			}
		}
		return result;
	}
};

int main(){
	vector<int> src;
	int input;
	int target;
	int i =0;

	vector<int> result;

	do{
		cin>>input;
		src.push_back(input);
		if(cin.get() == '
'){
			break;
		}
	}while(1);

	cin>>target;

	Solution s;
	result = s.twoSum(src, target);

	for(i=0; i<result.size(); i++){
		cout<<result[i]<<endl;
	}

	return 0;
}