[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]]

[LeetCode]–31. Next Permutation

做过这个题之后，肯定启发很多，所以我很快就写出来了。

public List<List<Integer>> permute(int[] nums) {        List<List<Integer>> res = new ArrayList<List<Integer>>();        if (nums.length == 0)            return res;        Arrays.sort(nums);        joinlist(nums, res);        findPermute(nums, res);        return res;    }    private void findPermute(int[] nums, List<List<Integer>> res) {        int i = nums.length - 1;        while (i > 0 && nums[i] <= nums[i - 1])            i--;        if (i == 0)            return;        int second = Integer.MAX_VALUE, secondIndex = Integer.MAX_VALUE;        for (int j = nums.length - 1; j > i - 1; j--)             if (nums[j] > nums[i - 1] && nums[j] < second){                second = nums[j];                secondIndex = j;            }        int temp = nums[i - 1];        nums[i - 1] = nums[secondIndex];        nums[secondIndex] = temp;        Arrays.sort(nums, i, nums.length);        joinlist(nums, res);        findPermute(nums, res);    }    private void joinlist(int[] nums, List<List<Integer>> res) {        List<Integer> list = new ArrayList<Integer>();        for (int i = 0; i < nums.length; i++)            list.add(nums[i]);        res.add(list);    }

找了一个另外的答案，也AC了，但是效率没有我写的高，不过看起来代码简洁一些。可以参考参考。简单理解一下就是找出规律，总共有n！这么多个序列，可以分为(n-1)!这么多个组，每个组n个，这样来通过for循环找到所有答案。

public List<List<Integer>> permute(int[] nums) {         ArrayList<List<Integer>> rst = new ArrayList<List<Integer>>();         if (nums == null) {             return rst;          }         if (nums.length == 0) {            rst.add(new ArrayList<Integer>());            return rst;         }         ArrayList<Integer> list = new ArrayList<Integer>();         helper(rst, list, nums);         return rst;    }    public void helper(ArrayList<List<Integer>> rst, ArrayList<Integer> list, int[] nums){        if(list.size() == nums.length) {            rst.add(new ArrayList<Integer>(list));            return;        }        for(int i = 0; i < nums.length; i++){            if(list.contains(nums[i])){                continue;            }            list.add(nums[i]);            helper(rst, list, nums);            list.remove(list.size() - 1);        }    }