LeetCode的medium題集合(C++實現)五

標籤：遞迴   c++   leetcode   

1 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.
可以採用遞迴的方法解決這個問題，當找到一組數之和等於目標後將這組數加入容器，然後返回；當一組數之和大於目標，立即返回；當小於目標繼續遞迴。

void dfscombine(vector<int>& candidates,int level,int& sum,int target,vector<int>& mid,vector<vector<int> >& result)    {        if(sum>target) return;        else if(sum==target)        {           result.push_back(mid);           return;        }        else        {          for(int i=level;i<candidates.size();i++)          {              sum+=candidates[i];              mid.push_back(candidates[i]);              dfscombine(candidates,i,sum,target,mid,result);              mid.pop_back();              sum-=candidates[i];          }        }    }    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {        vector<int> mid;          vector<vector<int> > result;          sort(candidates.begin(),candidates.end());        int level=0,sum=0;        dfscombine(candidates,level,sum,target,mid,result);        return result;    }

2 Combination Sum II
Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. Each number in C may only be used once in the combination.
該題與上一題的區別是給定的資料集中的資料只能用一次，我只需將上題中遞迴的參數level=i 改為 level=i+1 即可。同時，與前面講過的3sum等問題類似，防止從容器中pop_back() 出來的數與即將加入容器中的數相等而造成重複。

void dfscombine(vector<int>& candidates,int level,int& sum,int target,vector<int>& mid,vector<vector<int> >& result)    {        if(sum>target) return;        else if(sum==target)           result.push_back(mid);        else        {          for(int i=level;i<candidates.size();i++)          {              sum+=candidates[i];              mid.push_back(candidates[i]);              dfscombine(candidates,i+1,sum,target,mid,result); //保證一個元素只用一次              mid.pop_back();              sum-=candidates[i];              while(i<candidates.size()-1 && candidates[i]== candidates[i+1]) i++;  //防止重複          }        }    }    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {        vector<int> mid;          vector<vector<int> > result;          sort(candidates.begin(),candidates.end());        int level=0,sum=0;        dfscombine(candidates,level,sum,target,mid,result);        return result;    }

LeetCode的medium題集合(C++實現)五