2 sum, 3 sum, 3 sum close

1. The Sum of the

Given an array of integers, return indices of the both numbers such that they add-to a specific target.

You may assume this each input would has exactly one solution.

Example:

Given nums = [2, 7, one, 2], target = 9,because nums[0] + nums[1] = + 7 = 9,return [0, 1].

 Public classSolution { Public int[] Twosum (int[] Nums,inttarget) {        int[] res =New int[2]; if(Nums = =NULL|| Nums.length = = 0)returnRes; HashMap<integer, integer> map =NewHashmap<>();  for(inti = 0; i < nums.length; i++){            if(Map.containskey (target-Nums[i])) {res[0] = Map.get (target-Nums[i]); res[1] =i;        } map.put (Nums[i], i); }        returnRes; }}

3 Sum

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: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]]

 Public classSolution { PublicList<list<integer>> Threesum (int[] nums) {List<List<Integer>> res =NewArraylist<>();        Arrays.sort (Nums);  for(inti = 0; i < nums.length-2; i++) {List<Integer> list =NewArraylist<>(); intleft = i + 1; intright = Nums.length-1; intsum = 0-Nums[i];  while(Left <Right ) {                if(Nums[left] + nums[right] = =sum)                    {List.add (nums[i]);                    List.add (Nums[left]);                    List.add (Nums[right]); if(!res.contains (list)) Res.add (NewArraylist<>(list));                    List.clear ();  while(Left < right && nums[left] = = nums[left+1]) left++;  while(Left < right && nums[right] = = Nums[right-1]) right--; Left++; Right--; }                Else if(Nums[left] + Nums[right] <sum) { Left++; }                Else Right--; }        }        returnRes; }}

3Sum Closest

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); intMin =Integer.max_value;  for(inti = 0; i < nums.length-2; i++){          intleft = i + 1; intright = Nums.length-1; intValue = target-Nums[i];  while(Left <Right ) {                if(Nums[left] + nums[right] = =value)returnTarget; if(min = = Integer.max_value | | Math.Abs (Nums[i] + nums[left] + nums[right]-target) < Math.Abs (Min-target)) Min= Nums[i] + nums[left] +Nums[right]; if(Nums[left] + nums[right] >value) right--; Else Left++ ; }            }                returnmin; }}

2 sum, 3 sum, 3 sum close