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）

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;
            }
        }
    }
}

 Li Shi的 iterative 做法（还未深究）

public class Solution {
    public List<List<Integer>> permute(int[] num) {
        int len = num.length;
        List<List<Integer>> resSet = new ArrayList<List<Integer>>();
        List<Integer> curSeq = new ArrayList<Integer>();
        boolean[] used = new boolean[len];
        Arrays.fill(used,false);
        Arrays.sort(num);
        int curIndex = 0;
        curSeq.add(-1);
        while (curIndex!=-1){
            if (curIndex==len){
                List<Integer> temp = new ArrayList<Integer>();
                for (int i=0;i<curSeq.size();i++) temp.add(num[curSeq.get(i)]);
                resSet.add(temp);
                curIndex--;
                continue;
            }
            
            int val = -1;
            if (curSeq.size()>curIndex){
                val = curSeq.get(curIndex);
                if (val!=-1)
                    used[val]=false;
                curSeq.remove(curIndex);
            }

            boolean find = false;
            for (int i=val+1;i<len;i++)
                if (!used[i]){
                    used[i]=true;
                    curSeq.add(i);
                    curIndex++;
                    find = true;
                    break;
                }
            
            if (find)
                continue;
            else 
                curIndex--;
        }

        return resSet;
    }
}