Permutations 2
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],
[2,1,1]
]
Solution
public class Solution {
public List<List<Integer>> permuteUnique(int[] nums) {
List<List<Integer>> res = new ArrayList<>();
if(nums == null || nums.length == 0) return res;
boolean [] used = new boolean[nums.length];
List<Integer> list = new ArrayList<>();
Arrays.sort(nums);
dfs(nums, used, list, res);
return res;
}
public void dfs(int [] nums, boolean [] used, List<Integer> list, List<List<Integer>> res) {
if(list.size() == nums.length) {
res.add(new ArrayList<>(list));
return;
}
for(int i = 0; i< nums.length;i++) {
if(used[i]) continue;
if(i > 0 && nums[i] == nums[i-1] && !used[i - 1]) continue;
list.add(nums[i]);
used[i] = true;
dfs(nums, used, list, res);
used[i] = false;
list.remove(list.size() -1);
}
}
}
public class Solution {
public List<List<Integer>> permuteUnique(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
Arrays.sort(nums);
dfs(result,new ArrayList<>(), nums, new boolean[nums.length]);
return result;
}
public void dfs(List<List<Integer>> result, List<Integer> path, int [] nums, boolean [] visited) {
if(path.size() == nums.length) {
result.add(new ArrayList<>(path));
return;
} else {
for(int i = 0;i<nums.length;i++) {
if(visited[i]) continue;
if(i>0 &&nums[i-1]==nums[i] && !visited[i-1]) continue;
path.add(nums[i]);
visited[i] = true;
dfs(result, path, nums, visited);
path.remove(path.size() - 1);
visited[i] = false;
}
}
}
}