Code & Func

第19天

这道题是在起床到去上课前AC出来的，emmm，大概就10多分钟的样子。。。

虽然后来尝试优化了一下，但是感觉效果都不怎么好。。

题目描述：

Given a set of candidate numbers (C) (without duplicates) 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:

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[
2
  [7],
3
  [2, 2, 3]
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]

其实想法很简单，我既然想求combinationSum(7)，通过遍历数组，我们现在有了一个[2]，我只需要在求combinatiomSum(7-2)即可，然后组合起来：

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vector<int> cand;
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vector<vector<int> > ret;
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vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
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    cand = candidates;
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    sort(cand.begin(),cand.end());
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    vector<int> now;
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    combinationSumIter(now,0,target);
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    return ret;
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}
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void combinationSumIter(vector<int> &now,int beg,int target){
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    //cout << "target" << target << endl;
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    for(int i = beg;i < cand.size();++i) {
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        if (target < cand[i])
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            break;
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        else if (target == cand[i]) {
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            vector<int> vec = now;
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            vec.push_back(cand[i]);
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            ret.push_back(vec);
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        } else if (target - cand[i] >= cand[0] ){
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            vector<int> vec = now;
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            vec.push_back(cand[i]);
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            combinationSumIter(vec,i,target-cand[i]);
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        }
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    }
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}

然后在dicuss中看到的也是类似的想法：

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class Solution {
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public:
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    std::vector<std::vector<int> > combinationSum(std::vector<int> &candidates, int target) {
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        std::sort(candidates.begin(), candidates.end());
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        std::vector<std::vector<int> > res;
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        std::vector<int> combination;
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        combinationSum(candidates, target, res, combination, 0);
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        return res;
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    }
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private:
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    void combinationSum(std::vector<int> &candidates, int target, std::vector<std::vector<int> > &res, std::vector<int> &combination, int begin) {
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        if (!target) {
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            res.push_back(combination);
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            return;
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        }
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        for (int i = begin; i != candidates.size() && target >= candidates[i]; ++i) {
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            combination.push_back(candidates[i]);
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            combinationSum(candidates, target - candidates[i], res, combination, i);
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            combination.pop_back();
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        }
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    }
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};