4Sum

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 ≤ b ≤ c ≤ d)
• The solution set must not contain duplicate quadruplets.

 solution: use recursion method.

    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)

class Solution {public:    vector<vector<int>> ThreeSum(int start, vector<int>& nums, int target){        vector<vector<int>> result;          for(int a = start; a <= nums.size() - 3;)          {              int b = a + 1;              int c = nums.size() - 1;              while(b < c)              {                  int sum = nums[a] + nums[b] + nums[c];                  if(sum == target){                      result.push_back({nums[start-1], nums[a], nums[b], nums[c]});                      c--;                      while(b < c && nums[c] == nums[c+1]) c--;                      b++;                      while(b < c && nums[b] == nums[b-1]) b++;                  }                  else if(sum > target){                      c--;                      while(b < c && nums[c] == nums[c+1]) c--;                  }else{                      b++;                      while(b < c && nums[b] == nums[b-1]) b++;                  }              }              a++;              while(a <= nums.size() - 3 && nums[a] == nums[a-1]) a++;          }          return result;      }    vector<vector<int>> fourSum(vector<int>& nums, int target) {       if(nums.size() < 4){           return {};       }       sort(nums.begin(), nums.end());       vector<vector<int>> result;       for(int i = 0 ; i < nums.size() - 3;)       {           vector<vector<int>> vectors = ThreeSum(i+1, nums, target - nums[i]);           if(vectors.size() > 0)                result.insert(result.end(), vectors.begin(), vectors.end());            i++;            while(i < nums.size() - 3 && nums[i] == nums[i - 1]) i++;       }       return result;    }};