算法题：2sum，3sum

Answer 1

参考回答：

2sum：

vector<int> twoSum(vector<int>& nums, int target) {
    unordered_map<int,vector<int>> mark;
    for(int i=0;i<nums.size();i++)
        mark[nums[i]].push_back(i);
        for(int i = 0;i<nums.size();i++){
            if(target-nums[i] == nums[i]){
            if(mark[nums[i]].size() > 1){
           		 vector<int> tmp{i,mark[nums[i]][1]};
            	return tmp;
            }
            }else{
            if(mark.find(target-nums[i]) != mark.end()){
           		 vector<int> tmp{i,mark[target-nums[i]][0]};
 				return tmp;
            }
        }
    }
}

3sum：

void two_sum(vector<int>& nums,int i,int target,vector<vector<int>> &result){
    int j = nums.size()-1;
    int b = i-1;
    while(i<j){
        if(nums[i]+nums[j] == target){
            result.push_back(vector<int>{nums[b],nums[i],nums[j]});
            i++;
            j--;
            while(i<j && nums[i] == nums[i-1]) i++;
            while(i<j && nums[j+1] == nums[j]) j--;
        }else{
            if(nums[i]+nums[j] < target)
                i++;
            else
              j--;
        }
    }
    return;
}
vector<vector<int>> threeSum(vector<int>& nums) {
    set<vector<int>> result;
    vector<vector<int>> result2;
    sort(nums.begin(),nums.end());
    for(int i=0;i<nums.size();i++)
   		 if(i>0&&nums[i-1]== nums[i])
   		 continue;
    else
   		 two_sum(nums,i+1,-nums[i],result2);
    return result2;
}