47- Permutations II
Total Accepted: 75673 Total Submissions: 264907 Difficulty: Medium
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]
]
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public class Solution {
public List<List<Integer>> permuteUnique(int[] nums) {
/*https://leetcode.com/discuss/73856/really-easy-solution-easier-than-solutions-with-very-high-vote
Use an extra boolean array " boolean[] used" to indicate whether the value is added to list.
Sort the array "int[] nums" to make sure we can skip the same value.
when a number has the same value with its previous, we can use this number only if his previous is used*/
List<List<Integer>> res = new ArrayList<>();
List<Integer> cur = new ArrayList<>();
if (nums == null || nums.length == 0) {
return res;
}
boolean[] visited = new boolean[nums.length];
Arrays.sort(nums); //
backtrack(nums, visited, cur, res);
return res;
}
public void backtrack(int[] nums, boolean[] visited, List<Integer> cur, List<List<Integer>> res) {
if (cur.size() == nums.length ) {
res.add( new ArrayList<Integer>(cur));
return;
}
for (int i = 0; i < nums.length; i++) {
if (!visited[i]) {
if (i > 0 && nums[i] == nums[i-1] && visited[i-1]) {
return;
}
cur.add(nums[i]);
visited[i] = true;
backtrack(nums, visited, cur, res);
visited[i] = false;
cur.remove(cur.size() - 1);
}
}
}
}
