Given a collection of distinct integers, return all possible permutations.

1. count==n, add tinylist to list;

2. iterate through the input array, if the int hasn‘t been added to the tinylist, add it to the tinylist, and continue dbp

class Solution {
    public List<List<Integer>> permute(int
 
[] nums) {
        List<List<Integer>> list=new ArrayList<>();
        boolean[] bool=new boolean[nums.length];
        findPermute(nums, list, new ArrayList<Integer>(), bool);
        return list;
    }
    public void findPermute(int[] nums, List<List<Integer>> list, List<Integer> tinyList, boolean
 
[] bool){
        if(tinyList.size()==nums.length){
            list.add(new ArrayList(tinyList));
            return;
        } 
        for(int i=0;i<nums.length;i++){
            if(!bool[i]){
                bool[i]=true;
                tinyList.add(nums[i]);
                findPermute(nums, list, tinyList, bool);
                bool[i]
 
=false;
                tinyList.remove(tinyList.size()-1);
            }
        }
    }
}

可以不用bool去存儲tinylist裏是否包含該數字，直接判斷tinylist.contains即可，就是會稍微比用bool存慢一點點，時間換空間

class Solution {
    public List<List<Integer>> permute(int[] nums) {
        List<List<Integer>> list=new ArrayList<>();
        findPermute(nums, list, new ArrayList<Integer>());
        return list;
    }
    public void findPermute(int[] nums, List<List<Integer>> list, List<Integer> tinyList){
        if(tinyList.size()==nums.length){
            list.add(new ArrayList(tinyList));
            return;
        } 
        for(int i=0;i<nums.length;i++){
            if(!tinyList.contains(nums[i])){
                tinyList.add(nums[i]);
                findPermute(nums, list, tinyList);
                tinyList.remove(tinyList.size()-1);
            }
        }
    }
}

註意，在back tracking的時候每次添加tinylist的時候都要新創建一個list加進去，要不然最後加進去的全是空的list！！（

list.add(new ArrayList(tinyList));）

leetcode 46-Permutation(medium)