LeetCode 114 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]

分析：

看到返回所有，应该想到回溯，即DFS，DFS就要注意恢复现场。

public class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if(candidates==null || candidates.length==0)
            return res;
        //排序保证结果按递增排列
        Arrays.sort(candidates);
        dfs(candidates, 0, target, new ArrayList<Integer>(), res);
        return res;
    }
    
    public void dfs(int[] candidates, int start, int target, List<Integer> item, List<List<Integer>> res){
        //递归终结
        if(target <= 0){
            if(target==0)
                res.add(new ArrayList<Integer>(item));
            return;
        }
        
        for(int i=start; i<candidates.length; i++){
            //因为i-1位置的元素已经用过多次，如果相等的话，i位置就不应该再处理
            if(i>0 && candidates[i]==candidates[i-1])
                continue;
            item.add(candidates[i]);
            //i没有增加，保证i位置的元素可以用多次，当target减小
            dfs(candidates, i, target-candidates[i], item, res);
            item.remove(item.size()-1);
        }
    }
}