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package practice.recursion;
import java.util.ArrayList;
import java.util.List;
/**
* Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order.
Example 1:
Input: nums = [1,2,3]
Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
Example 2:
Input: nums = [0,1]
Output: [[0,1],[1,0]]
Example 3:
Input: nums = [1]
Output: [[1]]
Constraints:
1 <= nums.length <= 6
-10 <= nums[i] <= 10
All the integers of nums are unique.
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/permutations
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
*/
public class Permutations {
public List<List<Integer>> s2(int[] nums) {
int len = nums.length;
List<List<Integer>> res = new ArrayList<>();
if (len == 0) return res;
boolean[] used = new boolean[len];
List<Integer> path = new ArrayList<>();
dfs(nums, 0, path, used, res);
return res;
}
private void dfs(int[] nums, int depth,
List<Integer> path, boolean[] used,
List<List<Integer>> res) {
int len = nums.length;
if (depth == len) {
res.add(new ArrayList<>(path));
return;
}
for (int i = 0; i < len; i++) {
if (!used[i]) {
path.add(nums[i]);
used[i] = true;
dfs(nums, depth + 1, path, used, res); // drill down
used[i] = false; // reverse
path.remove(path.size() - 1);
}
}
}
public List<List<Integer>> s1(int[] nums) {
List<List<Integer>> resList = new ArrayList<>();
List<Integer> res = new ArrayList<>();
dfs1(nums, res, resList);
return resList;
}
public void dfs1(int[] nums, List<Integer> res, List<List<Integer>> resList) {
// terminator
if (res.size() == nums.length) {
resList.add(new ArrayList<>(res));
return ;
}
for (int i = 0; i < nums.length; i++)
if (!res.contains(nums[i])) {
res.add(nums[i]);
dfs1(nums, res, resList); // drill down
res.remove(res.size() - 1); // reverse
}
}
}