LEET code -- Permutations

 Given a collection of distinct numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:
[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].


印出所有的排列組合，使樣回朔法來跑

用一個boo;陣列來記錄那些數字用過了
接著丟給下一個去紀錄

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class Solution { public: void find(vector<vector<int>> &ans, vector<int> &current, vector<int> &nums, bool *use){ if(current.size() == nums.size()){ ans.push_back(current); return ; } for(int i=0;i<nums.size();i++){ if(use[i]==false){ current.push_back(nums[i]); use[i]=true; find(ans,current,nums,use); current.pop_back(); use[i]=false; } } } vector<vector<int>> permute(vector<int>& nums) { int size=nums.size(); bool use[size]={0}; vector<vector<int>> ans; vector<int> current; find(ans,current,nums,use);//not &use //for(int i=0;i<size;i++)use[i]=false; return ans; } };

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旋轉的方式

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class Solution { public: vector<vector<int>> permute(vector<int>& nums) { vector<vector<int>> res; create(nums,res,0); return res; } void create(vector<int>& nums,vector<vector<int>> &res,int n) { if(n==nums.size()-1) res.push_back(nums); else { for(int i=n;i<nums.size();i++) { swap(nums[i],nums[n]); create(nums,res,n+1); swap(nums[i],nums[n]); } } } void swap(int &a,int &b) { int tmp=a; a=b; b=tmp; } };