subset 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],
  []
]

排序两个字提前写在试卷上
for循环里面都是对nums[i]进行操作的

class Solution {
    public List<List<Integer>> subsets(int[] nums) {
        //cc
        List<List<Integer>> result = new ArrayList<>();
        List<Integer> temp = new ArrayList<>();
        
        if (nums == null) return null;
        
        if (nums.length == 0) return result;
        
        //排序总是可以想一想的
        Arrays.sort(nums);
        backtrace(nums, 0, temp, result);
        
        return result;
    }
    
    public void backtrace(int[] nums, int start, List<Integer> temp, List<List<Integer>> result) {
        result.add(new ArrayList(temp));
        
        for (int i = start; i < nums.length; i++) {
            temp.add(nums[i]);
            //这里应该是i + 1
            backtrace(nums, i + 1, temp, result);
            temp.remove(temp.size() - 1);
        }
    }
}