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PermutationOfUniqueElements.java
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PermutationOfUniqueElements.java
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package Leetcode;
import java.util.ArrayList;
import java.util.List;
/**
* @author kalpak
*
* Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order.
*
* Example 1:
* Input: nums = [1,2,3]
* Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
*
* Example 2:
* Input: nums = [0,1]
* Output: [[0,1],[1,0]]
*
* Example 3:
* Input: nums = [1]
* Output: [[1]]
*
*
* Constraints:
*
* 1 <= nums.length <= 6
* -10 <= nums[i] <= 10
* All the integers of nums are unique.
*
*/
public class PermutationOfUniqueElements {
public static List<List<Integer>> permute(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
List<Integer> temp = new ArrayList<>();
backtrack(result, temp, nums);
return result;
}
private static void backtrack(List<List<Integer>> result, List<Integer> temp, int[] nums) {
// Base Case
if(temp.size() == nums.length) {
result.add(new ArrayList<>(temp));
return;
}
for(int i = 0; i < nums.length; i++) {
if(temp.contains(nums[i]))
continue;
temp.add(nums[i]);
// go down the recursion tree and explore further potential candidates.
backtrack(result, temp, nums);
// Remove the last entered number before going up the recursive tree
temp.remove(temp.size() - 1);
}
}
public static void main(String[] args) {
int[] nums = new int[]{1, 3 ,5};
System.out.println(permute(nums));
}
}