11/5 <backtracking> pseudo back BFS +

78. Subsets

While iterating through all numbers, for each new number, we can either pick it or not pick it
1, if pick, just add current number to every existing subset.
2, if not pick, just leave all existing subsets as they are.
We just combine both into our result.

 

For example, {1,2,3} intially we have an emtpy set as result [ [ ] ]
Considering 1, if not use it, still [ ], if use 1, add it to [ ], so we have [1] now
Combine them, now we have [ [ ], [1] ] as all possible subset

 

Next considering 2, if not use it, we still have [ [ ], [1] ], if use 2, just add 2 to each previous subset, we have [2], [1,2]
Combine them, now we have [ [ ], [1], [2], [1,2] ]

 

Next considering 3, if not use it, we still have [ [ ], [1], [2], [1,2] ], if use 3, just add 3 to each previous subset, we have [ [3], [1,3], [2,3], [1,2,3] ]
Combine them, now we have [ [ ], [1], [2], [1,2], [3], [1,3], [2,3], [1,2,3] ]

class Solution {
    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        result.add(new ArrayList<>());
        for(int n : nums){
            int size = result.size();
            for(int i = 0; i < size; i++){
                List<Integer> subset = new ArrayList<>(result.get(i));
                subset.add(n);
                result.add(subset);
            }
        }
        return result;
    }
}

90. Subsets II

Before we use on a digital recording lastAdded to processing, and then determines whether the current number is the same as above and, if different, the number of cycles from 0 to the current or a subset, if the same, then the new set of subtracting the number of sub when the cycle number of subsets as a starting point for cycle

class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        Arrays.sort(nums);
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        result.add(new ArrayList<Integer>());
        int lastAdded = 0;
        for(int i = 0; i < nums.length; i++){
            if(i == 0 || nums[i] != nums[i-1]) lastAdded = 0;
            int size = result.size();
            for(int j = lastAdded; j < size; j++){
                List<Integer> cur = new ArrayList<Integer>(result.get(j));
                cur.add(nums[i]);
                result.add(cur);
            }
            lastAdded = size;
        }
        return result;
    }
}