lintcode:Permutations

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Given a list of numbers, return all possible permutations.

Challenge
Do it without recursion.

1.递归

class Solution {public:    /**     * @param nums: A list of integers.     * @return: A list of permutations.     */    vector<vector<int> > res;    void dfs(vector<int> &nums,vector<int> &sub,int cur){        if(cur==nums.size()){            res.push_back(sub);            return;        }        for(int i=0;i<nums.size();i++){            bool ok=true;            for(int j=0;j<cur;j++){                if(nums[i]==sub[j]){                    ok=false;                }            }            if(ok){                sub[cur]=nums[i];                dfs(nums,sub,cur+1);            }        }    }    vector<vector<int> > permute(vector<int> nums) {        // write your code here        int len=nums.size();        if(len==0){            return res;        }        vector<int> sub(len);        dfs(nums,sub,0);        return res;    }};

2.非递归

关于求下一个排列见以前的一篇博文 lintcode：Next Permutation

class Solution {public:    /**     * @param nums: A list of integers.     * @return: A list of permutations.     */    void nextPermutation(vector<int> &nums){        int len=nums.size();        if(len==1){            return;        }        int partitionIndex;        int in=len-1;        while(in){            if(nums[in-1]<nums[in]){                partitionIndex=in-1;                break;            }            in--;        }        //注意这里是in==0而不是partitionIndex==0;        //如果in==0，如果nums整个都是逆序的        if(in==0){            reverse(nums.begin(),nums.end());        }else{            int changeIndex;            for(int i=len-1;i>partitionIndex;i--){                if(nums[i]>nums[partitionIndex]){                   changeIndex=i;                   break;                }            }            swap(nums[partitionIndex],nums[changeIndex]);            reverse(nums.begin()+partitionIndex+1,nums.end());        }    }    int factorial(int n){        return (n==1||n==0)?1:factorial(n-1)*n;    }    vector<vector<int> > permute(vector<int> nums) {        // write your code here        vector<vector<int> > res;        if(nums.size()==0){            return res;        }        int sum=factorial(nums.size());        res.push_back(nums);        for(int i=1;i<sum;i++){            nextPermutation(nums);            res.push_back(nums);        }        return res;    }};

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