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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0001_0100.s0090_subsets_ii;

// #Medium #Array #Bit_Manipulation #Backtracking #Algorithm_II_Day_9_Recursion_Backtracking
// #2022_06_20_Time_2_ms_(82.94%)_Space_43.5_MB_(77.86%)

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

/**
 * 90 - Subsets II\.
 *
 * Medium
 *
 * Given an integer array `nums` that may contain duplicates, return _all possible subsets (the power set)_.
 *
 * The solution set **must not** contain duplicate subsets. Return the solution in **any order**.
 *
 * **Example 1:**
 *
 * **Input:** nums = [1,2,2]
 *
 * **Output:** [ [],[1],[1,2],[1,2,2],[2],[2,2]] 
 *
 * **Example 2:**
 *
 * **Input:** nums = [0]
 *
 * **Output:** [ [],[0]] 
 *
 * **Constraints:**
 *
 * *   `1 <= nums.length <= 10`
 * *   `-10 <= nums[i] <= 10`
**/
@SuppressWarnings("java:S5413")
public class Solution {
    List> allComb = new ArrayList<>();
    List comb = new ArrayList<>();
    int[] nums;

    public List> subsetsWithDup(int[] nums) {
        Arrays.sort(nums);
        this.nums = nums;
        dfs(0);
        allComb.add(new ArrayList<>());
        return allComb;
    }

    private void dfs(int start) {
        if (start > nums.length) {
            return;
        }
        for (int i = start; i < nums.length; i++) {
            if (i > start && nums[i] == nums[i - 1]) {
                continue;
            }
            comb.add(nums[i]);
            allComb.add(new ArrayList<>(comb));
            dfs(i + 1);
            comb.remove(comb.size() - 1);
        }
    }
}