4sum

k-sum问题都可以有2-sum引申出来解决，但是时间复杂度为O(n^k-1)。应该要用哈希解决才好的，之后再看看

class Solution{private:  vector<vector<int> > ans;public:  vector<vector<int> >fourSum(vector<int> &num, int target){    sort(num.begin(),num.end());    ans.clear();    for(int i = 0; i < num.size(); ++i) {      if(i > 0 && num[i] == num[i-1]) continue;      for(int j = i+1; j < num.size(); ++j){          if(j > i+1 && num[j] == num[j-1]) continue;          int k = j+1;          int t = num.size()-1;          while(k < t){            if(k > j+1 && num[k] == num[k-1]){              k++;              continue;            }            if(t < num.size()-1 && num[t] == num[t+1]){              t--;              continue;            }            int sum = num[i]+num[j]+num[k]+num[t];            if(sum == target){              vector<int>a;              a.push_back(num[i]);              a.push_back(num[j]);              a.push_back(num[k]);              a.push_back(num[t]);              ans.push_back(a);              k++;            }            else if(sum < target)              k++;            else               t--;          }       }    }    return ans;  }};