leetcode78. Subsets

题目要求

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

Note: 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],
  []
]

类似的题目有：
leetcode60 Permutation Sequence 可以参考这篇博客
leetcode77 Combinations 可以参考这篇博客

思路一：递归

还是利用递归的方式，在前一种情况的基础上遍历下一轮的组合情况。

    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        result.add(new ArrayList<Integer>());
        subsets(result, nums, 0, new ArrayList<Integer>());
        return result;
    }
    
    public void subsets(List<List<Integer>> result, int[] nums, int startIndex, List<Integer> currentList){
        if(startIndex == nums.length){
            return;
        }
        while(startIndex<nums.length){
            currentList.add(nums[startIndex++]);
            result.add(new ArrayList<Integer>(currentList));
            subsets(result, nums, startIndex, currentList);
            currentList.remove(currentList.size()-1);
        }
    }

思路2：排序后循环

起始subset集为：[]
添加S0后为：[], [S0]
添加S1后为：[], [S0], [S1], [S0, S1]
添加S2后为：[], [S0], [S1], [S0, S1], [S2], [S0, S2], [S1, S2], [S0, S1, S2]

public List<List<Integer>> subsets(int[] S) {
    List<List<Integer>> res = new ArrayList<>();
    res.add(new ArrayList<Integer>());
    
    for(int i : S) {
        List<List<Integer>> tmp = new ArrayList<>();
        for(List<Integer> sub : res) {
            List<Integer> a = new ArrayList<>(sub);
            a.add(i);
            tmp.add(a);
        }
        res.addAll(tmp);
    }
    return res;
}

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