[leetcode] Subsets II

Given a collection of integers that might contain duplicates, S, return all possible subsets.

Note:

• Elements in a subset must be in non-descending order.
• The solution set must not contain duplicate subsets.

https://oj.leetcode.com/problems/subsets-ii/

 

思路：关键是去重，先排序，然后根据后一个是否跟前一个相等来判断是否继续递归。

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

public class Solution {
    public ArrayList<ArrayList<Integer>> subsetsWithDup(int[] num) {

        ArrayList<ArrayList<Integer>> ans = new ArrayList<ArrayList<Integer>>();
        ArrayList<Integer> tmp = new ArrayList<Integer>();
        // sort first
        Arrays.sort(num);
        sub(num, 0, tmp, ans);
        return ans;
    }

    public void sub(int[] num, int k, ArrayList<Integer> tmp, ArrayList<ArrayList<Integer>> ans) {
        ArrayList<Integer> arr = new ArrayList<Integer>(tmp);
        ans.add(arr);

        for (int i = k; i < num.length; i++) {
            // remove the dup
            if (i != k && num[i] == num[i - 1])
                continue;

            tmp.add(num[i]);
            sub(num, i + 1, tmp, ans);
            tmp.remove(tmp.size() - 1);
        }
    }

    public static void main(String[] args) {
        System.out.println(new Solution().subsetsWithDup(new int[] { 1, 2, 2 }));
    }
}

 

第二遍记录：

只用增量法实现了去重。

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

public class Solution {
    public ArrayList<ArrayList<Integer>> subsetsWithDup(int[] num) {

        ArrayList<ArrayList<Integer>> ans = new ArrayList<ArrayList<Integer>>();
        ArrayList<Integer> tmp = new ArrayList<Integer>();
        // sort first
        Arrays.sort(num);
        sub(num, 0, tmp, ans);
        return ans;
    }

    public void sub(int[] num, int k, ArrayList<Integer> tmp, ArrayList<ArrayList<Integer>> ans) {
        ArrayList<Integer> arr = new ArrayList<Integer>(tmp);
        ans.add(arr);

        for (int i = k; i < num.length; i++) {

            if (i == k || num[i] != num[i - 1]) {
                tmp.add(num[i]);
                sub(num, i + 1, tmp, ans);
                tmp.remove(tmp.size() - 1);
            }
        }
    }

    public static void main(String[] args) {
        System.out.println(new Solution().subsetsWithDup(new int[] { 1, 2, 2 }));
    }
}

 

cc150上也看到了，第三遍走起。

　　注意先排序。

　　注意递归条件中的start，表明可以开始选取元素的索引，因为我们每次只允许从后面选，递归向下传递start值为 i+1（不是start+1,开始这里搞错了）。

import java.util.ArrayList;

public class Solution {
    public static ArrayList<ArrayList<Integer>> subsetsWithDup(int[] num) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        if (num == null || num.length == 0)
            return res;
        ArrayList<Integer> tmp = new ArrayList<Integer>();
        subsetsHelper(num, res, tmp, 0);

        return res;
    }

    private static void subsetsHelper(int[] num, ArrayList<ArrayList<Integer>> res, ArrayList<Integer> tmp, int start) {
        res.add(new ArrayList<Integer>(tmp));

        for (int i = start; i < num.length; i++) {

            if (i == start || num[i] != num[i - 1]) {
                tmp.add(num[i]);
                subsetsHelper(num, res, tmp, i + 1);
                tmp.remove(tmp.size() - 1);
            }
        }

    }

    public static void main(String[] args) {
        int[] a = { 1, 2, 2 };

        System.out.println(subsetsWithDup(a));
    }
}

 

 

 

 

参考：

http://www.cnblogs.com/lautsie/p/3249869.html