## 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
return;
}
for (int i = start; i < nums.length; i++) { // 扩展状态
if (gap < nums[i]) return; // 剪枝