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.

 

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)

思路I: 用求3Sum的方法,外加一次for循环,时间复杂度O(n3)

class Solution {public:    vector
     
      
       > fourSum(vector
       
        & nums, int target) {        int size = nums.size();        if(size < 3) return result;                sort(nums.begin(), nums.end());        for(int i = 0; i < size-3; i++){            if(i > 0 && nums[i]==nums[i-1])  continue; //The solution set must not contain duplicate=>no duplicate in the same position            for(int j = i+1; j < size-2; j++){                if(j> i+1 && nums[j]==nums[j-1]) continue; //The solution set must not contain duplicate=>no duplicate in the same position                find(nums, j+1, size-1, target-nums[i]-nums[j], i, j);            }        }        return result;    }    void find(vector
        
         & nums, int start, int end, int target, int& index1, int& index2){        int sum;        while(start
         
          no duplicate in the same position                    start++;                }while(start!= end && nums[start] == nums[start-1]);                do{ //The solution set must not contain duplicate=>no duplicate in the same position                    end--;                }while(end!=start && nums[end] == nums[end+1]);            }            else if(sum>target){                do{ //The solution set must not contain duplicate=>no duplicate in the same position                    end--;                }while(end!=start && nums[end] == nums[end+1]);            }            else{
   //The solution set must not contain duplicate=>no duplicate in the same position                do{                     start++;                }while(start!= end && nums[start] == nums[start-1]);            }        }    }    private:    vector
          
           
            > result; vector
            
              item;};
            
           
          
         
        
       
      
     

 

思路II：用hash table。O(N^2)把所有pair存入hash表，pair中两个元素的和就是hash值。那么接下来求4sum就变成了在所有的pair value中求 2sum，这个就成了线性算法了。所以整体上这个算法是O(N^2)+O(n) = O(N^2)。

class Solution {public:    vector
     
      
       > fourSum(vector
       
        & nums, int target) {        int size = nums.size();        int a, b, c, d;        vector
        
         
          > result;        unordered_map
          
           
            
              > > mp; unordered_map
             
               cnt; //各个数的数量 if(size < 3) return result; sort(nums.begin(), nums.end()); for(int i = 0; i < size-1; i++){ if(i > 0 && nums[i]==nums[i-1]) continue; for(int j = i+1; j < size; j++){ if(j> i+1 && nums[j]==nums[j-1]) continue; mp[nums[i]+nums[j]].push_back(pair
              
               {nums[i],nums[j]}); } } for(int i = 0; i < size; i++){ cnt[nums[i]]++; } for(unordered_map
               
                
                 
                   > >::iterator it1=mp.begin();it1!=mp.end();it1++){ //遍历map unordered_map
                  
                   
                    
                      > >::iterator it2=mp.find(target - it1->first); //查找map if(it2==mp.end()) continue;// not found if(it1->first > it2->first) continue; //already checked，去重 for(int i = 0; i < it1->second.size(); i++){ //访问map元素 for(int j = 0; j < it2->second.size(); j++){ a = it1->second[i].first; //访问pair元素 b = it1->second[i].second; c = it2->second[j].first; d = it2->second[j].second; if(max(a,b) > min(c,d)) continue; //四个数两两组合，有6种情况，这里只取两个最小的数在it1的情况，去重 cnt[a]--; cnt[b]--; cnt[c]--; cnt[d]--; if(cnt[a]<0||cnt[b]<0||cnt[c]<0||cnt[d]<0){ cnt[a]++; cnt[b]++; cnt[c]++; cnt[d]++; continue; } cnt[a]++; cnt[b]++; cnt[c]++; cnt[d]++; vector
                     
                       tmp = {a,b,c,d}; sort(tmp.begin(),tmp.end()); result.push_back(tmp); } } } return result; }};