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# 40 Combination Sum II

## 40. [Combination Sum II](https://leetcode.com/problems/combination-sum-ii/description/)

## 1. Question

Given a collection of candidate numbers (**C**) and a target number (**T**), find all unique combinations in **C** where the candidate numbers sums to **T**.

Each number in **C** may only be used **once** in the combination.

**Note:**

* All numbers (including target) will be positive integers.
* The solution set must not contain duplicate combinations.

For example, given candidate set`[10, 1, 2, 7, 6, 1, 5]`and target`8`,\
A solution set is:

```
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]
```

## 2. Implementation

**(1) Backtracking**

```java
class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<>();
        List<Integer> combinations = new ArrayList<>();
        Arrays.sort(candidates);
        getCombinations(candidates, 0, 0, target, combinations, res);
        return res;
    }

    public void getCombinations(int[] candidates, int index, int sum, int target, List<Integer> combinations, List<List<Integer>> res) {
        if (sum > target) {
            return;
        }

        if (sum == target) {
            res.add(new ArrayList<>(combinations));
            return;
        }

        for (int i = index; i < candidates.length; i++) {
            if (i > index && candidates[i - 1] == candidates[i]) {
                continue;
            }

            combinations.add(candidates[i]);
            getCombinations(candidates, i + 1, sum + candidates[i], target, combinations, res);
            combinations.remove(combinations.size() - 1);
        }
    }
}
```

## 3. Time & Space Complexity

**Backtracking**: 时间复杂度O(2^n), 空间复杂度O(2^n)