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Copy path_47.java
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60 lines (52 loc) · 1.96 KB
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package com.fishercoder.solutions;
import com.fishercoder.common.utils.CommonUtils;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
/**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]
]*/
public class _47 {
/**credit: https://discuss.leetcode.com/topic/31445/really-easy-java-solution-much-easier-than-the-solutions-with-very-high-vote*/
public List<List<Integer>> permuteUnique(int[] nums) {
List<List<Integer>> result = new ArrayList();
if (nums == null || nums.length == 0) return result;
boolean[] used = new boolean[nums.length];
List<Integer> list = new ArrayList();
Arrays.sort(nums);
dfs(nums, used, list, result);
return result;
}
private void dfs(int[] nums, boolean[] used, List<Integer> list, List<List<Integer>> result) {
if (list.size() == nums.length){
result.add(new ArrayList(list));
return;
}
for (int i = 0; i < nums.length; i++){
if (used[i]) continue;
if (i > 0 && nums[i - 1] == nums[i] && !used[i - 1])
continue;
/**
* For this line, both !used[i-1] and used[i-1] will AC. It is because the first one makes sure when
* duplicates are selected, the order is ascending (index from small to large). However,
* the second one means the descending order.
*/
used[i] = true;
list.add(nums[i]);
dfs(nums, used, list, result);
used[i] = false;
list.remove(list.size()-1);
}
}
public static void main(String...args){
int[] nums = new int[]{1,1,2};
_47 test = new _47();
List<List<Integer>> result = test.permuteUnique(nums);
CommonUtils.printListList(result);
}
}