LeetCode 39. Combination Sum

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.
• 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]
] 

Subscribe to see which companies asked this question

 

用递归做该题显得复杂度很高, 但是我觉得没有什么更好的办法. 用递归最关键的是明确每一步所需要完成的操作, 这样写起来会比较容易.

 

class Solution {
public:
    void solve(int depth, int maxdepth, int target, vector<vector<int> >& result,vector<int>& candidates, vector<int>& ret)
    {
        for(int i=0;i<=target/candidates[depth];i++){
            if(target-i*candidates[depth]<0)break;
            ret[depth]=i;

            if(target-i*candidates[depth]==0){//完成目标,记录下来
                vector<int> tmp;
                tmp.clear();
                for(int i=0;i<=maxdepth;i++)
                    for(int j=0;j<ret[i];j++){
                        tmp.push_back(candidates[i]);
                    }
                result.push_back(tmp);
                return;
            }
            
            if(depth+1==maxdepth+1){//最后一个数字的处理
                continue;
            }

            solve(depth+1,maxdepth,target-i*candidates[depth],result,candidates,ret);
        }
    }

    vector<vector<int> > combinationSum(vector<int>& candidates, int target) {
        vector<vector<int> > result;
        sort(candidates.begin(),candidates.end());
        vector<int> ret;
        ret.resize(candidates.size()+1);
        solve(0,candidates.size()-1,target,result,candidates,ret);

        return result;
    }
};