LeetCode 47. Permutations II (Java版; Meidum)-蒲公英云

LeetCode 47. Permutations II (Java版; Meidum)

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LeetCode 47. Permutations II (Java版; Meidum)

题目描述

Given a collection of numbers that might contain duplicates, return all possible unique permutations.
Example:
Input: [1,1,2]
Output:
[
  [1,1,2],
  [1,2,1],
  [2,1,1]
]

第一次做; 核心: 1) 递归函数逻辑: 处理完[0,i], 当前处理索引i处的元素; [i,n-1]范围上的元素都没用过; HashSet去重, 防止索引i处使用重复的元素

class Solution { 
    public List<List<Integer>> permuteUnique(int[] nums) { 
        List<List<Integer>> list = new ArrayList<>();
        List<Integer> al = new ArrayList<>();
        core(list, al, nums, 0);
        return list;
    }
    //递归函数逻辑: [0,i]上的结果
    private void core(List<List<Integer>> list, List<Integer> al, int[] nums, int i){ 
        //base case
        if(i==nums.length){ 
            list.add(new ArrayList<Integer>(al));
            return;
        }
        //
        int index=i;
        //记录当前位置出现过的数字
        HashSet<Integer> set = new HashSet<>();
        for(int k=i; k<nums.length; k++){ 
            int cur = nums[k];
            if(set.contains(cur)){ 
                continue;
            }
            swap(nums, index, k);
            al.add(cur);
            set.add(cur);
            core(list, al, nums, i+1);
            //
            al.remove(al.size()-1);
            swap(nums, index, k);
        }
    }
    private void swap(int[] arr, int i, int j){ 
        int tmp = arr[i];
        arr[i] = arr[j];
        arr[j] = tmp;
    }
}