Leetcode: Permutations

Given a collection of 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], and [3,2,1].

Analysis: 这道题跟N-Queens，Sudoku Solver，Combination Sum，Combinations, Subsets, Generate Parentheses等一样，也是一个NP问题。

又是把所有满足条件的结果存到一个ArrayList里面， 之前也有类似的如Letter combination of a Phone Number这道题。这种类型的题其实形成了一个套路，套路就是，recursion参数包括最终结果的集合（ArrayList），input（String），递归层次level（int），某一条具体的路径Path，当前外部环境

一般来说是：ArrayList<ArrayList<>>, ArrayList<>, input, 当前外部环境（可以是sum remain, 可以如本例是visited过的数组元素），递归层次level（int）

Notice: 我在permute函数里面第9行到第16行的循环其实是不需要的, 而且使用boolean数组来记录访问某个节点与否似乎比添加删除arraylist要容易

 1 public class Solution {

 2     public ArrayList<ArrayList<Integer>> permute(int[] num) {

 3         ArrayList<ArrayList<Integer>> permutations = new ArrayList<ArrayList<Integer>>();

 4         ArrayList<Integer> permutation = new ArrayList<Integer>();

 5         ArrayList<Integer> collection = new ArrayList<Integer>();

 6         for (int each : num) {

 7             collection.add(each);

 8         }

 9         for (int i = 0; i < collection.size(); i++) {

10             int current = collection.get(i);

11             collection.remove(i);

12             permutation.add(current);

13             helper(permutations, permutation, current, collection);

14             permutation.remove(0);

15             collection.add(i, current);

16         }

17         return permutations;

18     }

19     

20     public void helper(ArrayList<ArrayList<Integer>> permutations, ArrayList<Integer> permutation, int current, ArrayList<Integer> collection) {

21         if (collection.isEmpty()) {

22             permutations.add(new ArrayList<Integer>(permutation));

23             return;

24         }

25         for (int j = 0; j < collection.size(); j++) {

26             current = collection.get(j);

27             collection.remove(j);

28             permutation.add(current);

29             helper(permutations, permutation, current, collection);

30             permutation.remove(permutation.size()-1);

31             collection.add(j, current);

32         }

33     }

34 }

第二遍的做法：

public class Solution {

    public ArrayList<ArrayList<Integer>> permute(int[] num) {

        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();

        ArrayList<Integer> element = new ArrayList<Integer>();

        boolean[] visited = new boolean[num.length];

        helper(num, result, element, visited);

        return result;

    }

    

    public void helper(int[] num, ArrayList<ArrayList<Integer>> result, ArrayList<Integer> element, boolean[] visited){

        if (element.size() == num.length){

            // duplicate element and add it to result (element would be changed from time to time. If directly use element

            // only result would be changed when element changed)

            result.add(new ArrayList<Integer>(element)); 

            return;

        }

        

        for(int i=0; i<num.length; i++){

            if(!visited[i]){

                visited[i] = true;

                element.add(num[i]);

                helper(num, result, element, visited);

                

                // After providing a complete permutation, pull out the last number, 

                element.remove(element.size()-1);

                visited[i] = false;

            }

        }

    }

}