DFS解法的两道题 Leetcode 46 Permutations & Leetcode 78 Subset

Leetcode 78 Subset

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],
  []
]

题目解析：这种返回all possible的题就可以用bfs的方法来解啦~思路是先把array排序(因为题里要求是non-descending)然后思路是从num里面依次读取数字，存到ArrayList<Integer>的list里面, 然后再把list存到ArrayList<ArrayList<Integer>>的res里顺序是这样的[], [1], [1, 2], [1, 2, 3], [2], [2, 3], [3]记得边界检测

 1 public class Solution {
 2     public static List<List<Integer>> subsets(int[] S) {
 3         List<List<Integer>> res = new ArrayList<List<Integer>>();
 4         if(S == null || S.length == 0)
 5             return res;
 6         Arrays.sort(S);
 7         ArrayList<Integer> list = new ArrayList<Integer>();
 8         dfs(res, list, S, 0);
 9         return res;
10     }
11     public static void dfs(List<List<Integer>> res, ArrayList<Integer> list, int[] S, int pos){
12         res.add(new ArrayList<Integer>(list));
13         for(int i = pos; i < S.length; i++){
14             list.add(S[i]);
15             dfs(res, list, S, i+1);
16             list.remove(list.size() - 1);
17         }
18     }
19 }

Leetcode 46 Permutations

Given a collection of numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:
[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].

题目解析：

这道题更是典型的DFS啦~直接用DFS模型就好了~至于检测某个数组的element是否被使用了这种情况就新建一个boolean的数组进行检测就好了~

public class Solution {
    public static List<List<Integer>> permute(int[] num) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        ArrayList<Integer> list = new ArrayList<Integer>();
        boolean[] flag = new boolean[num.length];
        dfs(res, list, num, flag);
        return res;
    }
    public static void dfs(List<List<Integer>> res, ArrayList<Integer> list, int[] num, boolean[] flag){
        if(list.size() == num.length){
            res.add(new ArrayList<Integer>(list));
            return;
        }
        for(int i = 0; i < num.length; i++){
            if(!flag[i]){
                flag[i] = true;
                list.add(num[i]);
                dfs(res, list, num, flag);
                list.remove(list.size() - 1);
                flag[i] = false;
            }
        }
    }
}