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]

题目大意：给定几个全为正数的集合和一个目标值，找到所有和等于目标值的数的组合，数可以重复，结果以递增序列返回，不能重复

题目难度：Medium

import java.util.*;

/**
 * Created by gzdaijie on 16/5/21
 * DFS 
 */
public class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        Arrays.sort(candidates);
        ArrayList tmp = new ArrayList();
        combination(candidates, target, result,tmp, 0, 0);
        return result;

    }

    private void combination(int[] candidates, int target, List<List<Integer>> result, ArrayList<Integer> tmp, int k, int sum) {
        if (sum == target) {
            result.add((ArrayList<Integer>) tmp.clone());
            return;
        }
        int len = candidates.length;
        for (int i = k; i < len; i++) {
            // 小于目标值时，从i开始继续尝试，因为可以重复
            if (sum + candidates[i] <= target) {
                tmp.add(candidates[i]);
                combination(candidates, target,result, tmp, i, sum + candidates[i]);
                tmp.remove(tmp.size() - 1);
            }
        }
    }
}