Leetcode 题解 - 搜索--Backtracking(19):子集合

[LeetCode] Subsets 子集合

 

Given a set of distinct integers, S, return all possible subsets.

Note:

• Elements in a subset must be in non-descending order.
• The solution set must not contain duplicate subsets.

 

For example,
If S = [1,2,3], a solution is:

[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]

 解题思路：

这道题目拿到手第一想法就是利用DFS，与以往不同的是子集的大小是从0到3，这个点需要额外注意，本初就是在主函数中构造一个for循环来解决这个问题，剩下的就是整张的dfs算法。当然我在此处天添加了visited 和 start，就我个人的经验来说，我认为i这种集合顺序性存在的题目，怎么说呢 就是 12345567  用第一个再用第二个再用第三个的 用start明显好一些。

class Solution {
    public List> subsets(int[] nums) {
        List> subset = new ArrayList<>();
        List subsetList = new ArrayList<>();
        boolean[] visited = new boolean[nums.length];
        for(int i=0; i <= nums.length; i++)
            help(subset, subsetList, visited, 0, i, nums);
        return subset;
    }
    private void help(List> subset, List subsetList,
                     boolean[] visited, int start, final int size, final int[] nums){
        if(subsetList.size() == size){
            subset.add(new ArrayList<>(subsetList));
            return;
        }
        
        for(int i = start; i < nums.length; i++){
            subsetList.add(nums[i]);
            help(subset, subsetList, visited, i + 1, size, nums);
            subsetList.remove(subsetList.size() - 1);
        } 
    }
}

class Solution {
    public List> subsets(int[] nums) {
        List> result = new ArrayList<>();
        List list = new ArrayList<>();
        result.add(new ArrayList());
        boolean[] visited = new boolean[nums.length];
        help(list, result, nums, 0, visited);        
        return result;
    }
    
    private void help(List list, List> result, int[] nums, int start,
                     boolean[] visited){
        for(int i = start; i < nums.length; i++){
            visited[i] = true;
            list.add(nums[i]);
            if(list.size() <= nums.length)
                result.add(new ArrayList(list));
            help(list, result, nums, i+1, visited);
            list.remove(list.size() - 1);
        }
        
    }
}