Two sum

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

//method1:
class Solution{
    public:
        vector<int> twoSum(vector<int>& nums, int target){
            unordered_map<int,int> m;
            vector<int> res;
            for(int i = 0; i < nums.size(); ++i){
                m[nums[i]] = i;
            }
            
            for(int i = 0; i < nums.size(); ++i){
                int t = target - nums[i];
                if(m.count(t) && m[t] != i){
                    res.push_back(i);
                    res.push_back(m[t]);
                    break;
                }
            }
            return res;
        }
};

//method2:
class Solution2{
    public:
        vector<int> twoSum(vector<int>& nums, int target){
            unordered_map<int,int> m;
            for(int i = 0; i < nums.size(); ++i){
                if(m.count(target - nums[i])){
                    return {i,m[target - nums[i]]};
                }
                m[nums[i]] = i;
            }
            return {};
        }
};


方法3：

#include <iostream>

#include <vector>

#include <map>

#include <unordered_map>

using namespace std;

class Solution{

public:

    vector<int> twoSum(vector<int>& nums, int target){

        vector<int> res;

        unordered_map<int,int> map;

        for(int i = 0; i < nums.size(); i++)

        {

            map[nums[i]] = i;

        }

        for(int i = 0; i < nums.size(); i++)

        {

            int left = target - nums[i];

            if(map.count(left) && i < map[left])

            {

                res.push_back(i);

                res.push_back(map[left]);

            }

        }

    }

};