[leetcode]78. Subsets数组子集

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],
  []
]

题意：

给定一个含不同整数的集合，返回其所有的子集

assumption:

1. do we need to return if subsets is empty?

2. what kind of order to return if there are many subsets

3. dupulicates in the given array?

solution:

1.

example:

nums = [1,   2,   3]

^

index

level0          [ ]

/  |   \

level1     [1]  [2]  [3]

/    \

level2 [1,2]   [1,3]

/

level3  [1,2,3]

level0:  add [] to result[ [] ],   use pointer: index to scan given array, add current element to path [1], pass [1] to next level,

level1:  add[1] to result[[][1]],  treat index element as a start, pick one in the remaining and added to the path [1,2], pass[1,2] to next level

level2: add[1,2] to result[[][1][1,2]] treat index element as a start, pick one in the remaining and added to the path [1,2, 3], pass[1,2,3] to next level

level3: add[1,2,3] to result[[][1][1,2][1,2,3]]

二、High Level带着面试官walk through：

生成ArrayList作为每条path的记录，扔到result里

以当前index为开头，call helper function, 使得在index之后剩下可用的item中选一个加到当前path后面

代码：

 class Solution {
     public List<List<Integer>> subsets(int[] nums) {
         List<List<Integer>> result = new ArrayList<>();
         List<Integer> path = new ArrayList<>();
         helper(nums,0, path, result);
         return result;
     }
     private void helper(int[]nums, int index, List<Integer> path, List<List<Integer>> result){
         result.add(new ArrayList<>(path));
 
         for(int i = index; i< nums.length; i++){
             path.add(nums[i]);
             helper(nums, i+1, path, result);
             path.remove(path.size()-1);
         }
     }
 }