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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0001_0100.s0047_permutations_ii;

// #Medium #Array #Backtracking #Algorithm_II_Day_10_Recursion_Backtracking
// #2023_08_11_Time_1_ms_(99.86%)_Space_44.4_MB_(45.65%)

import java.util.ArrayList;
import java.util.List;

/**
 * 47 - Permutations II\.
 *
 * Medium
 *
 * Given a collection of numbers, `nums`, that might contain duplicates, return _all possible unique permutations **in any order**._
 *
 * **Example 1:**
 *
 * **Input:** nums = [1,1,2]
 *
 * **Output:** [[1,1,2], [1,2,1], [2,1,1]] 
 *
 * **Example 2:**
 *
 * **Input:** nums = [1,2,3]
 *
 * **Output:** [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]] 
 *
 * **Constraints:**
 *
 * *   `1 <= nums.length <= 8`
 * *   `-10 <= nums[i] <= 10`
**/
public class Solution {
    private List> ans;

    public List> permuteUnique(int[] nums) {
        ans = new ArrayList<>();
        permute(nums, 0);
        return ans;
    }

    private void permute(int[] nums, int p) {
        if (p >= nums.length - 1) {
            List t = new ArrayList<>(nums.length);
            for (int n : nums) {
                t.add(n);
            }
            ans.add(t);
            return;
        }
        permute(nums, p + 1);
        boolean[] used = new boolean[30];
        for (int i = p + 1; i < nums.length; i++) {
            if (nums[i] != nums[p] && !used[10 + nums[i]]) {
                used[10 + nums[i]] = true;
                swap(nums, p, i);
                permute(nums, p + 1);
                swap(nums, p, i);
            }
        }
    }

    private void swap(int[] nums, int i, int j) {
        int t = nums[i];
        nums[i] = nums[j];
        nums[j] = t;
    }
}