"Leetcode" 18. 4Sum

Title Description:

Given an array S of n integers, is 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.

Problem Solving Analysis:

This and 3Sum the problem somewhat like, also must first determine two number, then two number of determination can refer to binary search implementation method to optimize.

Specific code:

1  Public classSolution {2      Public StaticList<list<integer>> Foursum (int[] Nums,inttarget) {3 Arrays.sort (nums);4         //the judgment of the boundary condition5list<list<integer>> results =NewArraylist<list<integer>>();6         if(nums.length<4){7             returnresults;8         }9         if(nums.length==4){Ten             if(nums[0]+nums[1]+nums[2]+nums[3]==target) { Onelist<integer> result =NewArraylist<integer>();  AResult.add (nums[0]); -Result.add (nums[1]); -Result.add (nums[2]); theResult.add (nums[3]); - Results.add (result); -  -             } +             returnresults; -         } +         //to determine the position of the first two numbers by ergodic method A          for(intfirst=0;first<nums.length-3;first++){ at              for(intsecond=first+1;second<nums.length-2;second++){ -                 //the third and fourth numbers are determined with a binary lookup -                 intThird=second+1; -                 intFourth=nums.length-1; -                  while(third<Fourth) { -                     if(nums[first]+nums[second]+nums[third]+nums[fourth]>target) { infourth--; -                     } to                     Else if(nums[first]+nums[second]+nums[third]+nums[fourth]<target) { +third++; -                     } the                     Else{ *list<integer> result =NewArraylist<integer>(); $ Result.add (Nums[first]);Panax Notoginseng Result.add (Nums[second]); - Result.add (Nums[third]); the Result.add (Nums[fourth]); + Results.add (result); A                         //Avoid element duplication the                          while(third<nums.length-1&&nums[third+1]==Nums[third]) { +third++; -                         } $third++; $                         //Avoid element duplication -                          while(third<fourth&&nums[fourth-1]==Nums[fourth]) { -fourth--; the                         } -fourth--;Wuyi                          the                     } -                 } Wu                 //Avoid element duplication -                  while(second<nums.length-2&&nums[second+1]==Nums[second]) { Aboutsecond++; $                 } -             } -             //Avoid element duplication -              while(first<nums.length-3&&nums[first+1]==Nums[first]) { Afirst++; +             } the         } -         returnresults; $     } the}

"Leetcode" 18. 4Sum