# Combination Sum

### 描述​

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 ($a_1, a_2, ..., a_k$) must be in non-descending order. (ie, $a_1 \leq a_2 \leq ... \leq a_k$).
• 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]

### 代码​

// Combination Sum// 时间复杂度O(n!)，空间复杂度O(n)public class Solution {    public List<List<Integer>> combinationSum(int[] nums, int target) {        Arrays.sort(nums);        List<List<Integer>> result = new ArrayList<>(); // 最终结果        List<Integer> path = new ArrayList<>(); // 中间结果        dfs(nums, path, result, target, 0);        return result;    }    private static void dfs(int[] nums, List<Integer> path,                            List<List<Integer>> result, int gap, int start) {        if (gap == 0) {  // 找到一个合法解            result.add(new ArrayList<Integer>(path));            return;        }        for (int i = start; i < nums.length; i++) { // 扩展状态            if (gap < nums[i]) return; // 剪枝            path.add(nums[i]); // 执行扩展动作            dfs(nums, path, result, gap - nums[i], i);            path.remove(path.size() - 1);  // 撤销动作        }    }}