11/5 <backtracking> 伪BFS+回溯

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

我们用 lastAdded 来记录上一个处理的数字，然后判定当前的数字和上面的是否相同，若不同，则循环还是从0到当前子集的个数，若相同，则新子集个数减去之前循环时子集的个数当做起点来循环

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;
    }
}