LeetCode 78. Subsets

原题链接在这里：https://leetcode.com/problems/subsets/

题目：

Given a set of distinct integers, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

Input: nums = [1,2,3]
Output:
[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]

题解：

Set state on each level. What are needed for the state. 

Here it needs to know index and current item.

Time Complexity: exponential.

Space: O(nums.length). stack space.

AC Java:

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

Iteration时, 取res中现有list，每个list都加新的元素nums[i]然后再放回res中，同时保留原有list. 从[]开始一次加一个新元素。

Note: 内层for loop 的size 一定要提前提取出来，不能在内层for中现提取，因为res的size一直在变.

Time Complexity: exponential. Space: O(res.size()).

 AC Java:

class Solution {
    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        res.add(new ArrayList<Integer>());
        for(int num : nums){
            //这里要注意，必须提前把size提取出来，不能把提取过程直接嵌入到下面的for语句中
            //因为res的size会在下面语句中一直变化
            int size = res.size();
            for(int i = 0; i<size; i++){
                List<Integer> copyItem = new ArrayList<Integer>(res.get(i));
                copyItem.add(num);
                res.add(copyItem);
            }
        }
        
        return res;
    }
}

类似Combinations, Combination Sum, Palindrome Partitioning.

进阶题是Subsets II.