leetcode -- 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, � , a_k) must be in non-descending order. (ie, a₁ ? a₂ ? � ? a_k).
• 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] 

本题是求解所有可能的组合数，可用DFS来求解，如需求解最优解，需使用DP

DFS的递归函数中只处理当前状态节点n，而不关心它的下一状态

public class Solution {
    public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int len = candidates.length;
        ArrayList<ArrayList<Integer>> results = new ArrayList<ArrayList<Integer>>();
        if(len == 0)
            return results;
        
        Arrays.sort(candidates);
        
        ArrayList<Integer> result = new ArrayList<Integer>();
        int level = 0;
        int sum = 0;
        DFS(candidates, level, sum, target, result, results);
        return results;
    }
    
    public void DFS(int[] candidates, int level, int sum, int target, ArrayList<Integer> result,
        ArrayList<ArrayList<Integer>> results){
        if(sum > target)
            return;
        
        if(sum == target){
            ArrayList<Integer> tmp = new ArrayList<Integer>();
            tmp.addAll(result);
            results.add(tmp);
            return;
        }
        if(sum < target){
            for(int i = level; i < candidates.length; i++){
                sum += candidates[i];
                result.add(candidates[i]);
                DFS(candidates, i, sum, target, result, results);
                result.remove(result.size() - 1);
                sum -= candidates[i];
            }
        }
    }
}

ref：

DFS 算法入门：http://rapheal.iteye.com/blog/1526863