lintcode:Permutations

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;
    }
};