[LeetCode] Combination Sum, Solution

最新推荐文章于 2019-05-10 10:55:45 发布

原创最新推荐文章于 2019-05-10 10:55:45 发布 · 226 阅读

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本博客探讨了在给定候选数集合中寻找所有可能的组合，这些组合的元素之和等于指定的目标值。允许重复选择候选数，且组合必须按非降序排列，确保结果集中不包含重复组合。

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 (a₁, a₂, â€¦ , a_k) must be in non-descending order. (ie, a₁ â‰¤ a₂ â‰¤ â€¦ â‰¤ 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]

» Solve this problem

[Thoughts]

This is a normal recursion question. For each candidate, add and verify the target. If it hit, add it as a part of solution.

[Code]

1:    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {  
2:      // Start typing your C/C++ solution below  
3:      // DO NOT write int main() function  
4:      vector<vector<int> > result;  
5:      vector<int> solution;  
6:      int sum=0;  
7:      std::sort(candidates.begin(), candidates.end());  
8:      GetCombinations(candidates,sum, 0, target, solution, result);  
9:      return result;  
10:    }  
11:    void GetCombinations(  
12:      vector<int>& candidates,  
13:      int& sum,  
14:      int level,  
15:      int target,  
16:      vector<int>& solution,  
17:      vector<vector<int> >& result)  
18:    {  
19:      if(sum > target) return;  
20:      if(sum == target)  
21:      {  
22:        result.push_back(solution);  
23:        return;  
24:      }  
25:      for(int i =level; i< candidates.size(); i++)  
26:      {  
27:        sum+=candidates[i];  
28:        solution.push_back(candidates[i]);  
29:        GetCombinations(candidates, sum, i, target, solution, result);  
30:        solution.pop_back();  
31:        sum-=candidates[i];        
32:      }  
33:    }