[Leetcode] Three sums && three cloest sums && four sums

Given an array S of n integers, is there elements a, b, c in S such That a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

• Elements in a triplet (a,b,C) must is in non-descending order. (ie, a ≤ b ≤ c)
• The solution set must not contain duplicate triplets.

    For example, given array S = {-1 0 1 2-1-4},    A solution set is:    ( -1, 0, 1)    (-1,-1, 2)


Idea: If the direct use of triple traversal brute force crack, the time complexity of O (n^3). It is possible to refer to the sum of two numbers, and for an ordered array, the time complexity of finding two numbers and N is O (n). Sorts the array, and the time complexity is O (NLog (n)). Finally, the problem becomes the number of the first selection, and then the remaining array of query Target-nums[i] Two number, pay attention to the method of checking the weight, then the total time complexity O (n^2) +o (Nlog (n)) =o (n^2).

 Public classSolution { PublicList<list<integer>> Threesum (int[] nums)        {Arrays.sort (nums); intlen=nums.length; List<List<Integer>> list=NewArraylist<>();  for(inti=0;i<len-2;i++)        {            //Go heavy            if(i>0 && nums[i]==nums[i-1])Continue; inttarget2=0-Nums[i]; Twosum (nums, I+1, Target2, list); }        returnlist; }     Public voidTwosum (int[] Nums,intStartintTarget,list<list<integer>>list) {        intHead=start, End=nums.length-1;  while(head<end) {            intsum=nums[head]+Nums[end]; if (sum<target) {head++;            } else if (sum>target) {end--; }            Else{List<Integer> temp=NewArraylist<>(); Temp.add (Nums[start-1]);                Temp.add (Nums[head]);                Temp.add (Nums[end]);                                List.add (temp); //eliminate the possibility of duplication, because it is an ordered array, from small to large, so as long as one one to find duplicates and skip can                intK=head+1;  while(K<end && Nums[k]==nums[head]) k++; Head=K; K=end-1;  while(K>head && nums[k]==nums[end]) k--; End=K; }        }    }}

Given an array S of n integers, find three integers in S such so the sum is closest to a give n number, target. Return the sum of the three integers. You may assume this each input would has exactly one solution.

    For example, given array S = {-1 2 1-4}, and target = 1.    The sum is closest to the target is 2. (-1 + 2 + 1 = 2).

 Public classSolution { Public intThreesumclosest (int[] Nums,inttarget)        {Arrays.sort (nums); intdis=Integer.max_value; intResult=0;  for(inti=0;i<nums.length-2;i++)        {            inttarget2=target-Nums[i]; intRe=twosumcloest (Nums, i+1, Target2); intD=math.abs (target-re-Nums[i]); if(d<dis) {Dis=D; Result=re+Nums[i]; if(d==0)                    returnTarget; }        }        returnresult; }     Public intTwosumcloest (int[] Nums,intStartinttarget) {        intHead=start,tail=nums.length-1; intRe=0,d=Integer.max_value;  while(head<tail) {            intsum=nums[head]+Nums[tail]; if(sum<target) {                intdis=target-sum; if(dis<d) {d=dis; Re=sum; } head++; }            Else if(target<sum) {                intdis=sum-Target; if(dis<d) {d=dis; Re=sum; } Tail--; }            Else {                returnTarget; }        }        returnre; }}

[Leetcode] Three sums && three cloest sums && four sums