这个解题思路来自代码随想录：代码随想录 (programmercarl.com)

class Solution {
public:vector<int> twoSum(vector<int>& nums, int target) {std::unordered_map <int,int> map;for(int i = 0; i < nums.size(); i++) {// 遍历当前元素，并在map中寻找是否有匹配的keyauto iter = map.find(target - nums[i]); if(iter != map.end()) {return {iter->second, i};}// 如果没找到匹配对，就把访问过的元素和下标加入到map中map.insert(pair<int, int>(nums[i], i)); }return {};}
};

 小编自己写的方法实测不佳，但可作为一个解题思路

// 方法一:效果差
class Solution {
public:vector<int> twoSum(vector<int>& nums, int target) {vector<int> v;int count = 0;unordered_map<int,pair<int,int>> mp;int tmp;for(int i=0;i<nums.size();i++) {count = 0;if(mp.count(target-nums[i])!=0) {count++;mp[target-nums[i]].second = count; }tmp = target-nums[i];mp.insert({nums[i],{tmp,count}});v.push_back(i);}for(int i=0;i<v.size();i++) {// cout<<"v[i]"<<v[i]<<endl;// cout<<"mp[nums[v[i]]].second"<<mp[nums[v[i]]].second<<endl;if(mp[nums[v[i]]].second==0) {v.erase(v.begin() + i);i--;}}return v;}
};// 方法二:效果不佳
class Solution {
public:vector<int> twoSum(vector<int>& nums, int target) {vector<int> v;int count = 0;unordered_map<int,pair<int,int>> mp;int tmp;int k=0;int tmp_index;for(int i=0;i<nums.size();i++) {tmp = target-nums[i];if(tmp==nums[i]) {k++;if(k==1) tmp_index = i;else if(k==2) {// mp.insert({nums[i],{tmp_index,i}});// k=0;return {tmp_index,i};}}else {if(mp.count(tmp)!=0) {mp.insert({nums[i],{-1,i}});}else{mp.insert({nums[i],{i,-1}});}}}for(auto it:mp) {if(it.second.second !=-1) {tmp = target-it.first;// cout<<it.first<<":"<<it.second.first<<","<<it.second.second<<endl;// cout<<mp.find(tmp)->first<<":"<<mp.find(tmp)->second.first<<","<<mp.find(tmp)->second.second<<endl;v.push_back(it.second.second);v.push_back(mp.find(tmp)->second.first);}}return v;}
};