Code & Fun

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

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

然后在dicuss中看到的也是类似的想法:

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

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