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LeetCode - Subsets

时间:2016-01-02 16:06:45      阅读:115      评论:0      收藏:0      [点我收藏+]

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题目:

Given a set of distinct integers, nums, 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 nums = [1,2,3], a solution is:

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

思路:

借用上一道题Combinations的代码

package recursion;

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

public class Subsets {

    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        List<Integer> record = new ArrayList<Integer>();
        Arrays.sort(nums);
        int n = nums.length;
        for (int k = 0; k <= n; ++k)
            generateRecord(res, record, nums, 0, n - 1, k);
        return res;
    }
    
    private void generateRecord(List<List<Integer>> res, List<Integer> record, int[] nums, int start, int end, int k) {
        if (k == 0) {
            res.add(record);
            return;
        }
        
        for (int i = start; i <= end - k + 1; ++i) { 
            List<Integer> newRecord = new ArrayList<Integer>(record);
            newRecord.add(nums[i]);
            generateRecord(res, newRecord, nums, i + 1, end, k - 1);
        }
    }
    
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Subsets s = new Subsets();
        int[] nums = { 1, 2, 3 };
        List<List<Integer>> res = s.subsets(nums);
        for (List<Integer> l : res) {
            for (int i : l) 
                System.out.print(i + "\t");
            System.out.println();
        }
    }

}

 

LeetCode - Subsets

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原文地址:http://www.cnblogs.com/shuaiwhu/p/5094671.html

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