## 原题

Given a set of candidate numbers (`candidates`(without duplicates) and a target number (`target`), find all unique combinations in `candidates` where the candidate numbers sums to `target`.

The same repeated number may be chosen from `candidates` unlimited number of times.

Note:

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

Example 1:

```Input: candidates = `[2,3,6,7], `target = `7`,
A solution set is:
[
[7],
[2,2,3]
]
```

Example 2:

```Input: candidates = [2,3,5]`, `target = 8,
A solution set is:
[
[2,2,2,2],
[2,3,3],
[3,5]
]```

## 讲义

回溯法（深度优先搜索，Depth-First Search）

## 代码

[code lang=”cpp”] class Solution { public: vector<vector <int>> combinationSum(vector<int>& candidates, int target) { vector<vector <int>> result{}; vector<int> candidate{}; sort(candidates.begin(), candidates.end()); sum(result, candidates, target, candidate, 0); return result; } void sum(vector<vector <int>>& result, vector<int>& candidates, int target, vector</int><int>& candidate, int level) { // up to the target if (target == 0) { result.push_back(candidate); return; } for (int i = level; i < candidates.size() && candidates[i] <= target; ++i) { candidate.push_back(candidates[i]); sum(result, candidates, target – candidates[i], candidate, i); candidate.pop_back(); } } }; [/code]

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