[leetcode] Subsets

Given a set of distinct integers, S, return all possible subsets.

Note:

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

For example,
If S = [1,2,3], a solution is:

[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]

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

思路：递归构造，每一层递归都选择填或者不填s[cur]的操作。

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

public class Solution {
    public ArrayList<ArrayList<Integer>> subsets(int[] S) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        ArrayList<Integer> tmp = new ArrayList<Integer>();
        Arrays.sort(S);
        subsetsHelper(res, tmp, S, 0);
        return res;
    }

    private void subsetsHelper(ArrayList<ArrayList<Integer>> res, ArrayList<Integer> tmp, int[] s, int cur) {
        if (cur == s.length) {
            res.add(new ArrayList<Integer>(tmp));
            return;
        }

        // not fill
        subsetsHelper(res, tmp, s, cur + 1);

        // fill
        tmp.add(s[cur]);
        subsetsHelper(res, tmp, s, cur + 1);
        tmp.remove(tmp.size() - 1);

    }

    public static void main(String[] args) {
        System.out.println(new Solution().subsets(new int[] { 4, 1, 0 }));
    }
}

第二遍记录：

子集一直不是很熟，重新复习了下。

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

public class Solution {
    public List<List<Integer>> subsets(int[] S) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if (S == null || S.length == 0)
            return res;
        Arrays.sort(S);
        List<Integer> tmp = new ArrayList<Integer>();
        subsetsHelper(res, tmp, S, 0);
        return res;
    }

    private void subsetsHelper(List<List<Integer>> res, List<Integer> tmp, int[] S, int cur) {
        if (cur == S.length) {
            res.add(new ArrayList<Integer>(tmp));
            return;
        } else {

            subsetsHelper(res, tmp, S, cur + 1);
            tmp.add(S[cur]);
            subsetsHelper(res, tmp, S, cur + 1);
            tmp.remove(tmp.size() - 1);

        }

    }

    public static void main(String[] args) {
        System.out.println(new Solution().subsets(new int[] { 4, 1, 0 }));
    }

}

C语言的 位向量法

#include<cstdio>
#include<cstdlib>
#include<ctime>
const int maxn = 50;
void print_subset(int n, int A[], int B[], int cur) {
    if (cur == n) {
        for (int i = 0; i < n; i++)
            if (B[i])
                printf("%d ", A[i]);
        printf("
");
        return;
    }
    B[cur] = 1; //choose element cur
    print_subset(n, A, B, cur + 1);
    B[cur] = 0; //not to choose element cur
    print_subset(n, A, B, cur + 1);
}
int main() {
    int n;
    int B[maxn] = { 0 };
    int A[] = { 1, 3, 5 };
    n=sizeof(A)/sizeof(int);
    print_subset(n,A, B, 0);
    return 0;
}

第四遍记录：

画递归调用图。

public class Solution {
    public List<List<Integer>> subsets(int[] s) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        List<Integer> tmp = new ArrayList<Integer>();
        Arrays.sort(s);
        subsetsHelper(res,tmp,s,0);
        return res;
    }
    
    private void subsetsHelper(List<List<Integer>> res, List<Integer> tmp, int[] s,int start){
           res.add(new ArrayList<Integer>(tmp));
           
           for(int i=start;i<s.length;i++){
                if(i==start||s[i]!=s[i-1]){
                    tmp.add(s[i]);
                    subsetsHelper(res,tmp,s,i+1);
                    tmp.remove(tmp.size()-1);
                }
           }
    }
}