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]
]

1.using extra space to check visited number:

public List<List<Integer>> permute(int[] nums) {        if(nums.length == 0)            return new ArrayList<List<Integer>>();        List<List<Integer>> result = new ArrayList<List<Integer>>();        Arrays.sort(nums);        List<Integer> list = new ArrayList<Integer>();        Set<Integer> set = new HashSet<Integer>();        helper(nums, list, result, set);        return result;    }    private void helper(int[] nums, List<Integer> list, List<List<Integer>> result, Set<Integer> set){        if(list.size() == nums.length){            List<Integer> newList = new ArrayList<Integer>(list);            result.add(newList);            return;        }        else{            for(int i= 0; i < nums.length; i++){                if(!set.contains(nums[i])){                    list.add(nums[i]);                    set.add(nums[i]);                    List<Integer> newList = new ArrayList<Integer>(list);                    helper(nums, newList, result, set);                    list.remove(list.size()-1);                    set.remove(nums[i]);                }            }        }    }

2.插入法
当n=1时，数组中只有一个数a1，其全排列只有一种，即为a1

当n=2时，数组中此时有a1a2，其全排列有两种，a1a2和a2a1，那么此时我们考虑和上面那种情况的关系，我们发现，其实就是在a1的前后两个位置分别加入了a2

当n=3时，数组中有a1a2a3，此时全排列有六种，分别为a1a2a3, a1a3a2, a2a1a3, a2a3a1, a3a1a2, 和 a3a2a1。那么根据上面的结论，实际上是在a1a2和a2a1的基础上在不同的位置上加入a3而得到的。

_ a1 _ a2 _ : a3a1a2, a1a3a2, a1a2a3

_ a2 _ a1 _ : a3a2a1, a2a3a1, a2a1a3

public List<List<Integer>> permute(int[] num) {    List<List<Integer>> result = new ArrayList<List<Integer>>();    List<Integer> list = new ArrayList<Integer>();    result.add(list);    for (int n : num) {        int size = result.size();        for (int i=0; i < size; i++) {            List<Integer> prevList = result.get(i);            for (int j = 0; j <= prevList.size(); j++) {                List<Integer> newList = new ArrayList<Integer>(prevList);                newList.add(j, n);                result.add(newList);            }        }    }    return res;}

3.交换法
in c++:

vector<vector<int> > permute(vector<int> &num) {        vector<vector<int> > res;        permuteDFS(num, 0, res);        return res;    }    void permuteDFS(vector<int> &num, int start, vector<vector<int> > &res) {        if (start >= num.size()) res.push_back(num);        for (int i = start; i < num.size(); ++i) {            swap(num[start], num[i]);            permuteDFS(num, start + 1, res);            swap(num[start], num[i]);        }    }