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+/*%0A39. Combination Sum My Submissions QuestionEditorial Solution%0ATotal Accepted: 95889 Total Submissions: 304005 Difficulty: Medium%0AGiven a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.%0A%0AThe same repeated number may be chosen from C unlimited number of times.%0A%0ANote:%0AAll numbers (including target) will be positive integers.%0AThe solution set must not contain duplicate combinations.%0AFor example, given candidate set %5B2, 3, 6, 7%5D and target 7, %0AA solution set is: %0A%5B%0A  %5B7%5D,%0A  %5B2, 2, 3%5D%0A%5D%0AHide Company Tags Snapchat Uber%0AHide Tags Array Backtracking%0AHide Similar Problems (M) Letter Combinations of a Phone Number (M) Combination Sum II (M) Combinations (M) Combination Sum III (M) Factor Combinations%0A*/%0Apublic class Solution %7B%0A    public List%3CList%3CInteger%3E%3E combinationSum(int%5B%5D candidates, int target) %7B%0A        List%3CList%3CInteger%3E%3E ans = new LinkedList%3CList%3CInteger%3E%3E();%0A        Arrays.sort(candidates);%0A        backtrack(ans,new LinkedList%3CInteger%3E(), candidates,target,0);%0A        return ans;%0A    %7D%0A    %0A    public void backtrack(List%3CList%3CInteger%3E%3E ans, List%3CInteger%3E temList, int%5B%5D candidates, int remain, int start)%7B%0A        if(remain%3C0)return;%0A        else if(remain==0)%7B%0A            ans.add(new LinkedList%3CInteger%3E(temList));%7D%0A        else%7B%0A            for(int i=start;i%3Ccandidates.length;i++)%7B%0A                temList.add(candidates%5Bi%5D);%0A                backtrack(ans,new LinkedList%3CInteger%3E(temList),candidates,remain-candidates%5Bi%5D,i);%0A                temList.remove(temList.size()-1);%0A            %7D%0A        %7D%0A    %7D%0A%7D%0A

Create Combination_Sum.java