LeetCode 39 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]

思路参考下图：

public List<List<Integer>> combinationSum(int[] candidates, int target) {
Arrays.sort(candidates);
List<List<Integer>> result = new LinkedList<List<Integer>>();
List<Integer> tmp = new ArrayList<Integer>();
dfs(0, candidates, target, tmp, result);
return result;
}

private void dfs(int begin, int[] candidates, int rest, List<Integer> tmp, List<List<Integer>> result) {
if (rest == 0)
result.add(new ArrayList<Integer>(tmp));//tmp是变动的,所以此处需要new一个
/*分别从candidates[i]开始往下深搜*/
for (int i = begin; i < candidates.length && candidates[i] <= rest; i++) {
tmp.add(candidates[i]);
dfs(i, candidates, rest - candidates[i], tmp, result);//搜索下个数字
tmp.remove(tmp.size() - 1);//回溯
}
}