LeetCode | 46. 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],  [3,2,1]]

题意：给出几个元素，生成这些元素的全排列。递归算法

//22 msclass Solution {public:    int vis[1000000];    vector<vector<int> > res;    void f(vector<int>& nums,vector<int>& tmp)    {        for(int i=0;i<nums.size();i++)        {            if(vis[i] == 0)         //还没有放进去，压入 vector            {                tmp.push_back(nums[i]);                vis[i] = 1;                f(nums,tmp);                vis[i] = 0;                vector<int>::iterator it = tmp.end();                it--;                tmp.erase(it);      //删除刚才放进去的元素            }        }        if(tmp.size() == nums.size())       //全部都放进去了        {            res.push_back(tmp);            return;        }    }    vector<vector<int>> permute(vector<int>& nums)    {        vector<int> tmp;        f(nums,tmp);        return res;    }};

解法二：solution提供的解法，每次直接对现有数组交换顺序，不需要额外空间，代码量也更少

// 13 msclass Solution {public:    vector<vector<int> > permute(vector<int> &num) {        vector<vector<int> > result;        permuteRecursive(num, 0, result);        return result;    }    // permute num[begin..end]    // invariant: num[0..begin-1] have been fixed/permuted    void permuteRecursive(vector<int> &num, int begin, vector<vector<int> > &result)    {        if (begin >= num.size()) {            // one permutation instance            result.push_back(num);            return;        }        for (int i = begin; i < num.size(); i++) {            swap(num[begin], num[i]);            permuteRecursive(num, begin + 1, result);            // reset            swap(num[begin], num[i]);        }    }};