LintCode : Permutations II

Posted 2020-06-14

import java.util.Collections;
class Solution {
    /**
     * @param nums: A list of integers.
     * @return: A list of unique permutations.
     */
    public ArrayList<ArrayList<Integer>> permuteUnique(ArrayList<Integer> nums){
        // write your code here
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if (nums == null || nums.size() == 0) {
            return result;
        }
        Collections.sort(nums);
        boolean[] visited = new boolean[nums.size()];
        ArrayList<Integer> list = new ArrayList<Integer>();
        permutationHelper(result, list, nums, visited);
        return result;
    }
    
    private void permutationHelper(ArrayList<ArrayList<Integer>> result, ArrayList<Integer> list, ArrayList<Integer> nums, boolean[] visited) {
        if (list.size() == nums.size()) {
            result.add(new ArrayList<Integer>(list));
            return;
        }
        
        for (int i = 0; i < nums.size(); i++) {
            if (visited[i] == true || (i != 0 && nums.get(i) == nums.get(i - 1) && visited[i - 1] == false)) {
                continue;
            }
            visited[i] = true;
            list.add(nums.get(i));
            permutationHelper(result, list, nums, visited);
            list.remove(list.size() - 1);
            visited[i] = false;
        }
    }
}

　　用visited数组来记录哪些是访问过的，visited[i - 1] == false很tricky，可能存在1222......这种情况，用第二个2递归的时候，前面那个2应该是还没有访问的，所以这个时候visited[i - 1] == false。