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# 40.Combination-Sum-II

## 40. Combination Sum II

## 题目地址

<https://leetcode.com/problems/combination-sum-ii/>

## 题目描述

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

Each number in candidates 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.

Example 1:
Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]
Example 2:

Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
  [1,2,2],
  [5]
]
```

## 代码

### Approach #1 Sort + Backtrace

`i != startIndex && candidates[i] == candidates[i - 1]` 去重

```java
class Solution {
        public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> results = new ArrayList<>();
        if (candidates == null || candidates.length == 0) {
            return results;
        }

        Arrays.sort(candidates);
        List<Integer> combination = new ArrayList<Integer>();
        dfs(candidates, 0, combination, target, results);

        return results;
    }

    private void dfs(int[] candidates, int startIndex, List<Integer> combination, int target, List<List<Integer>> results) {
        if (target == 0) {
            results.add(new ArrayList<Integer>(combination));
            return;
        }

        for (int i = startIndex; i < candidates.length; i++) {
            // IMPORTANT 去重
            if (i != startIndex && candidates[i] == candidates[i - 1]) {
                continue;
            }
            if (target < candidates[i]) {
                break;
            }
            combination.add(candidates[i]);
            dfs(candidates, i + 1, combination, target - candidates[i], results);
            combination.remove(combination.size() - 1);
        }
    } 

}
```


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