Leetcode Permutations II

題目地址：https://leetcode.com/problems/permutations-ii/

題目解析：和permutations同樣，只須要在交換元素是判斷元素是否重複便可

題目解答：

public class Solution {
    public List<ArrayList<Integer>> permuteUnique(int[] num) {
        List<ArrayList<Integer>> ret = new ArrayList<ArrayList<Integer>>();
        if(num == null || num.length == 0){
            return ret;
        }
        
        Arrays.sort(num); 
        permute(num,0,ret);
        return ret;
    }
    
    private void permute(int[] num,int start,List<ArrayList<Integer>> res){
        if(start == num.length - 1){
            ArrayList<Integer> temp = new ArrayList<Integer>();
            for(int i=0;i<num.length;i++){
                temp.add(num[i]);
            }
            
            res.add(temp);
            return;
        }
        
        for(int i=start;i<num.length;i++){
            if(containsDuplicate(num,start,i)){
                int temp = num[start];
                num[start] = num[i];
                num[i] = temp;
            
                permute(num,start+1,res);
            
                temp = num[start];
                num[start] = num[i];
                num[i] = temp;
            }
        }
    }
    
    private boolean containsDuplicate(int[] arr, int start, int end) {
        for (int i = start; i <= end-1; i++) {
            if (arr[i] == arr[end]) {
                return false;
            }
        }
        return true;
    }
}