Permutations II

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

For example,
[1,1,2] have the following unique permutations:
[1,1,2], [1,2,1], and [2,1,1].

和permutations 基本一致， 只是需要先把array 排序，然后保证在同一次循环中选取的数字不同即可。

public class Solution {
    ArrayList<ArrayList<Integer>> result = null;
    int len = 0;
    public ArrayList<ArrayList<Integer>> permuteUnique(int[] num) {
        // Start typing your Java solution below
        // DO NOT write main() function
        Arrays.sort(num);
        result = new ArrayList<ArrayList<Integer>>();
        if(num == null || num.length == 0) return result;
        len = num.length;
        boolean[] item = new boolean[len];
        get(item, num, new ArrayList<Integer>());
        return result;
    }
    public void get(boolean[] item, int[] num, ArrayList<Integer> row){
        if(row.size() == len){
            result.add(new ArrayList<Integer>(row));
            return;
        }
        for(int i = 0; i < len; i ++){
            if(!item[i]){
                item[i] = true;
                row.add(num[i]);
                get(item, num, row);
                item[i] = false;
                row.remove(row.size() - 1);
                while(i + 1 < num.length && num[i + 1] == num[i])  
                    i ++;
            }
        }
    }
}

第三遍： 不使用额外的boolean 数组：

public class Solution {
    public List<List<Integer>> permuteUnique(int[] num) {
        ArrayList<List<Integer>> result = new ArrayList<List<Integer>>();
        Arrays.sort(num);
        getPermutation(num, num.length, new ArrayList<Integer>(), result);
        return result;
    }
    public void getPermutation(int[] num, int len, ArrayList<Integer> row, ArrayList<List<Integer>> result){
        if(num == null || len == 0){
            result.add(row);
            return;
        }
        int pre = Integer.MAX_VALUE;
        for(int i = 0; i < num.length; i ++){
            if(num[i] != Integer.MAX_VALUE && num[i] != pre){
                pre = num[i];
                num[i] = Integer.MAX_VALUE;
                ArrayList<Integer> nrow = new ArrayList<Integer>(row);
                nrow.add(pre);
                getPermutation(num, len - 1, nrow, result);
                num[i] = pre;
            }
        }
    }
}

第四遍：

public class Solution {
    ArrayList<List<Integer>> result;
    public List<List<Integer>> permuteUnique(int[] num) {
        result = new ArrayList<List<Integer>>();
        if(num == null || num.length == 0) return result;
        Arrays.sort(num);
        findPermutation(num, new ArrayList<Integer>());
        return result;
    }
    
    private void findPermutation(int[] num, ArrayList<Integer> row){
        if(row.size() == num.length){
            result.add(row);
        }
        int pre = Integer.MIN_VALUE;
        for(int i = 0; i < num.length; i ++){
            if(num[i] != Integer.MIN_VALUE && num[i] != pre){
                pre = num[i];
                row.add(num[i]);
                num[i] = Integer.MIN_VALUE;
                findPermutation(num, new ArrayList<Integer>(row));
                num[i] = pre;
                row.remove(row.size() - 1);
            }
        }
    }
}