Combination Sum

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.

Note:

• 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.

For example, given candidate set 2,3,6,7 and target 7, 
A solution set is: 
[7] 
[2, 2, 3] 

思路

先对数组进行升序排列，然后在用回溯法BS，这里用的递归调用实现的

public class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        Arrays.sort(candidates);
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        getResult(result, new ArrayList<Integer>(), candidates, target, 0);
        
        return result;
    }
    
    private void getResult(List<List<Integer>> result, List<Integer> cur, int candidates[], int target, int start){
        if(target > 0){
            for(int i = start; i < candidates.length && target >= candidates[i]; i++){
                cur.add(candidates[i]);
                getResult(result, cur, candidates, target - candidates[i], i);
                cur.remove(cur.size() - 1);
            }//for
        }//if
        else if(target == 0 ){
            result.add(new ArrayList<Integer>(cur));
        }//else if
    }
}