39.Combination Sum

https://leetcode.com/problems/combination-sum/

1.Description(Medium)

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Notice

• All numbers (including target) will be positive integers.

• Elements in a combination (a1,a2, … ,ak) must be in non-descending order. (ie,a1≤a2≤ … ≤ak).

• The solution set must not contain duplicate combinations.

Example

Given candidate set[2,3,6,7]and target7, a solution set is:

[7]
[2, 2, 3]

2.Code

跟palindrome partition很像，经典DFS+backtracking

http://www.jiuzhang.com/solutions/combination-sum/

//如果candidates里面有重复的值，那么结果里面也会有重复的，所以要把candidates去重
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> result=new ArrayList<>();
        List<Integer> path=new ArrayList<Integer>();
        int[] nums=removeDuplicate(candidates);
        dfs(nums,0,target,path,result);
        return result;

    }

    public void dfs(int[] candidates,int startindex,int target,List<Integer> path,List<List<Integer>> result){
        if(target==0){
            result.add(new ArrayList<Integer>(path));
            return;
        }
        for(int i=startindex;i<candidates.length;i++){
            if(candidates[i]>target){
                break;
            }
            path.add(candidates[i]);
            //每次从i开始就能保证不会有重复的答案。
            dfs(candidates,i,target-candidates[i],path,result);
            path.remove(path.size()-1);
        }

    }

    //Remove duplicate for array(Leetcode problem)
    public int[] removeDuplicate(int[] candidates){
        HashSet<Integer> set=new HashSet<Integer>();
        for(int i=0;i<candidates.length;i++){
            set.add(candidates[i]);
        }

        int[] result=new int[set.size()];
        //Iterator<Integer> it=set.iterator();
        int index=0;
        //while(it.hasNext()){
        //    result[index]=it.next();
            //index++;
        //}

        for (int element : set){
            result[index++] = element;
        }
        Arrays.sort(result);
        return result;
    }

Version 2:

DFS+BackTracking

class Solution {
    private List<List<Integer>> result= new ArrayList<>();
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        if(candidates == null || candidates.length == 0) return result;
        ArrayList<Integer> path = new ArrayList<Integer>();
        dfs(candidates, 0, path, target);
        Set<List<Integer>> set = new HashSet<>();
        for(List<Integer> item: result)
        {
            Collections.sort(item);
            set.add(item);
        }
        return new ArrayList<>(set);
    }
    
    public void dfs(int[] candidates, int start, ArrayList<Integer> path, int target)
    {
       if(target == 0)
       {
           result.add(new ArrayList<Integer>(path));
       }
        for(int i = start; i< candidates.length; i++)
        {
            if(candidates[i] > target)
            {
                continue;
            }
            path.add(candidates[i]);
            dfs(candidates,i,path,target-candidates[i]);
            path.remove(path.size()-1);
        }
    }

}