4Sum Problem

4SumJan 27 '12

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

• Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a â‰¤ b â‰¤ c â‰¤ d)
• The solution set must not contain duplicate quadruplets.

    For example, given array S = {1 0 -1 0 -2 2}, and target = 0.    A solution set is:    (-1,  0, 0, 1)    (-2, -1, 1, 2)    (-2,  0, 0, 2)

方法1：利用和3Sum一样的思路，但是复杂度为O(N^3)，偷懒用之前的代码，堆在一起看起来就很屎，懒得改了。

代码：

class Solution {public:    vector<vector<int> > fourSum(vector<int> &num, int target) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        // Start typing your C/C++ solution below        // DO NOT write int main() function        sort(num.begin(),num.end());            vector<vector<int>> ret4;    set<vector<int>> set4;        //visit each positive number as target and use 2Sum to the left ones          int target3;          vector<int>::iterator it = num.begin();          for(;it!=num.end();){  if(it!=num.begin() && *it == *(it-1)) {it++;continue;}            target3 =target-*it;                //erase the element              it = num.erase(it);              //3Sum              set<vector<int>> set3 = threeSum(num,target3,target-target3);set4.insert(set3.begin(),set3.end());            //insert the element              it = num.insert(it,target-target3);              it++;          }          ret4.insert(ret4.end(),set4.begin(),set4.end());          return ret4;      }set<vector<int> > threeSum(vector<int> &num,int target3,int target) {        // Start typing your C/C++ solution below        // DO NOT write int main() functionvector<vector<int>> ret3;//visit each positive number as target and use 2Sum to the left onesint target2;set<vector<int>> set3;vector<int>::iterator it = num.begin();for(;it!=num.end();){if(it!=num.begin() && *it == *(it-1)) {it++;continue;}target2 = target3 -*it;//erase the elementit = num.erase(it);//2Sum2set<vector<int>> set2 = twoSum2(num,target2,target3,target);set3.insert(set2.begin(),set2.end());//insert the elementit = num.insert(it,target3-target2);it++;}return set3;    }     set<vector<int>> twoSum2(vector<int> &numbers, int target2,int target3,int target) {            // Start typing your C/C++ solution below             // DO NOT write int main() function                              //match             int i = 0;            int j = numbers.size()-1;            set<vector<int>> ret2;           while(i<j){                if(numbers[i]+numbers[j]==target2){                   vector<int> ret;                  ret.push_back(numbers[i]);                  ret.push_back(numbers[j]);                  ret.push_back(target3-target2); ret.push_back(target);sort(ret.begin(),ret.end());                ret2.insert(ret);                  i++;                  j--;              }else if(numbers[i]+numbers[j]>target2){                  j--;              }else{                  i++;              }              }            return ret2;       }};

时间：1800 milli secs

 

方法2： 对num生成两两的数对，然后对数对间进行2Sum，这样空间复杂度为O(N^2)，时间复杂度也为O(N^2)。

代码改天补上