leetcode_77_combiantion_78_subsets_90_subsets2

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真是有种 “x了狗了”的感觉，DFS这里总是充满了各种各样的问题

最后写的代码可读性和复杂度都差得厉害，虽然AC：

class Solution {public:void solve(int index,int cur,vector<int> &res,vector<vector<int> >&ans,vector<int>& num,int k){if(index==k){ans.push_back(res);return;}for(int i=cur;i<num.size();i++){res.push_back(num[i]);//index++;             //如果一开始按照index++的思路写，会影响for循环之后的多次递归，所以用index+1代替solve(index+1,i+1,res,ans,num,k); res.pop_back();}}    vector<vector<int> > combine(int n, int k) {        vector<vector<int> > ans;        vector<int> res;        vector<int> num;        for(int i=0;i<n;i++)        num.push_back(i+1);        solve(0,0,res,ans,num,k);        return ans;    }};

好吧，就在这篇文章之后继续做DFS，发现subsets和subsets2可以按照之前写的框架写：

传送门：

https://leetcode.com/problems/subsets/

其中每一次进行递归时都会产生新的集合，直接加进去就好

class Solution {public:void solve(int index,int cur,vector<int> &res,vector<vector<int> >&ans,vector<int>& num){ans.push_back(res);for(int i=cur;i<num.size();i++){res.push_back(num[i]);//index++;solve(index+1,i+1,res,ans,num); res.pop_back();}}    vector<vector<int> > subsets(vector<int> &S) {        vector<vector<int> > ans;        vector<int> res;        //vector<int> num;        sort(S.begin(),S.end());        solve(0,0,res,ans,S);        return ans;    }};

subsets2:

多加一个判定函数就好

传送门：https://leetcode.com/problems/subsets-ii/

AC代码：

class Solution {public:    bool check(vector<int> &s,vector<vector<int> >&ans)    {        for(int k=0;k<ans.size();k++)            if(ans[k]==s) return false;        return  true;    }void solve(int index,int cur,vector<int> &res,vector<vector<int> >&ans,vector<int>& num){if(check(res,ans))ans.push_back(res);for(int i=cur;i<num.size();i++){res.push_back(num[i]);//index++;solve(index+1,i+1,res,ans,num); res.pop_back();}}   vector<vector<int> > subsetsWithDup(vector<int> &S) {        vector<vector<int> > ans;        vector<int> res;        //vector<int> num;        sort(S.begin(),S.end());        solve(0,0,res,ans,S);        return ans;    }};

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