LeetCode90 subsets-ii

Title description

Given an integer set S that may contain repeated elements, return all subsets of the integer set.
Note:
The elements in the subset you give should be arranged in non-increasing order
. The solution set cannot contain repeated subsets.
For example:
if S = [1,2,2], the solution set should be:
[ ↵ [2], ↵ [1], ↵ [1,2,2], ↵ [2,2], ↵ [1,2], ↵ [] ↵]

analysis

  • It is very similar to LeetCode78 subsets, but it needs to consider the duplication.
  • Method 1: Pick the appropriate element at each position and use a backtracking algorithm.
  • Method 2, each element is in the corresponding position, there are options to join the set and not to join the set, using a backtracking algorithm. But the order is not guaranteed and fails.

Method 1 code

import java.util.*; 
public class Solution {
    ArrayList<ArrayList<Integer>> listAll = new ArrayList<>(); 
    public ArrayList<ArrayList<Integer>> subsetsWithDup(int[] num) {
        if (num == null || num.length <= 0)
            return listAll;
        ArrayList<Integer> list = new ArrayList<>();
        Arrays.sort(num); 
        Findsubset(num, 0, list);      
        return listAll;
    }
 
    public void Findsubset(int[] set, int start, ArrayList<Integer> list) {
        listAll.add(new ArrayList<>(list)); 
        //在某个位置选个一个数 从下标start开始选择
        for (int i = start; i < set.length; i++) {
            //i > start 重复序列 第一个可以放进去 其他不放
            //重复序列的首个加入 后面重复的才有机会加入
            //首个 不加入 后边都不加入
            //这样保证只考虑在此位置 只放首个元素,其他重复元素不放在此位置。后面的要加入,则在下一位置
            if (i > start && set[i] == set[i - 1])
                continue;
            list.add(set[i]);
            Findsubset(set, i + 1, list);
            list.remove(list.size() - 1);
        }
    }
}

Method 2

 public  ArrayList<ArrayList<Integer>> subsets(int[] S) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<>();
        Arrays.sort(S);
        helperDupli(S,0,true,new ArrayList<>(),res);
        return  res;
    }
private void helperDupli(int[] s, int curIndex,boolean taken, ArrayList<Integer> cur, ArrayList<ArrayList<Integer>> res) {
        if(curIndex == s.length){
            res.add(new ArrayList<>(cur));
        }else{

            //CurIndex处的 s元素不加入集合内
            helperSub(s,curIndex+1,cur,res);
            //taken 为true表示上一个curIndex处元素,加入了。此时,此处若是重复元素,他有加入的机会。不是重复的元素更有机会
            //taken false 表示上一个curIndex处元素,不加入。此时,此处若是重复元素,不能加入。即两个条件都不满足。。
            //若不是重复元素 可以加入
            if(taken ||s[curIndex-1]  != s[curIndex]){
                //curIndex 处 s元素加入集合内
                cur.add(s[curIndex]);
                helperSub(s,curIndex+1,cur,res);
                cur.remove(cur.size()-1);
            }
        }
    }
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Origin blog.csdn.net/weixin_40300702/article/details/105560996