DFS_46. 全排列

输入: [1,2,3]
输出:
[
[1,2,3],
[1,3,2],
[2,1,3],
[2,3,1],
[3,1,2],
[3,2,1]
]

class Solution {
  public static List<List<Integer>> permute(int[] nums) {
        List<List<Integer>> res = new LinkedList();
        if (nums.length == 0){
            return res;
        }
        int n = nums.length;
        Deque<Integer> path = new ArrayDeque<Integer>();
        boolean [] used = new boolean[nums.length];
        dfs(nums,0,path,used,res);
        return res;
    }

    private static void dfs(int[] nums, int depth, Deque<Integer> path, boolean[] used, List<List<Integer>> res) {
        //如果全部都判断完了就记录并返回
        if (depth == nums.length){
            res.add(new ArrayList<>(path));
            return;
        }
        for (int i = 0; i < nums.length; i++) {
            if (used[i]){
                continue;
            }
            path.add(nums[i]);
            used[i] = true;
            dfs(nums,depth + 1,path,used,res);
            //返回上一层的同时条件给他初始化回去
            path.remove(nums[i]);
            used[i] = false;
        }

    }
}

class Solution {
  public void backtrack(int n,ArrayList<Integer> output,List<List<Integer>> res,int first) {
    // 所有数都填完了
    if (first == n)
      res.add(new ArrayList<Integer>(output));
    for (int i = first; i < n; i++) {
      // 动态维护数组
      Collections.swap(output, first, i);
      // 继续递归填下一个数
      backtrack(n, output, res, first + 1);
      // 撤销操作
      Collections.swap(output, first, i);
    }
  }

  public List<List<Integer>> permute(int[] nums) {
    List<List<Integer>> res = new LinkedList();

    ArrayList<Integer> output = new ArrayList<Integer>();
    for (int num : nums)
      output.add(num);

    int n = nums.length;
    backtrack(n, output, res, 0);
    return res;
  }
}