Combinations

Link: https://leetcode.com/problems/combinations/

Constraint:

    1 <= n <= 20
    1 <= k <= n   ==> k is always valid

Idea

Initialize an structure to keep track of whether a number has been visited or not
Initialize a list to keep track of the combination visited so far
Search:
If the size of combinatin list is equal to k, add it to final result list
For each number i in range [1, N]:
if i is not visited:
add the number to the combination list, mark this number as visited
repeat this earch on other numbers recursively, with the starting index from (i + 1)
remove the number i from the combination, mark it in the as unvisited

Code

class Solution {
    public List<List<Integer>> combine(int n, int k) {
        List<List<Integer>> results = new ArrayList<>();
        search(n, k, 1, new boolean[n + 1], new ArrayList<Integer>(), results);
        return results;
    }
    
    private void search(int n, int k, int start, boolean[] visited, List<Integer> combination, List<List<Integer>> results) {
        if (combination.size() == k) {
            results.add(new ArrayList<>(combination));
            return;
        }
        
        for (int i = start; i <= n; i++) {
            if (!visited[i]) {
                visited[i] = true;
                combination.add(i);
                search(n, k, i + 1, visited, combination, results);
                combination.remove(combination.size() - 1);
                visited[i] = false;
            }
        }
    }
}

• Time: It would be equal to C(n, k), n choose k.
• Space:

Slightly better version without the need to use boolean array:

class Solution {
   public List<List<Integer>> combine(int n, int k) {
       List<List<Integer>> results = new ArrayList<>();
       search(n, k, 1, new ArrayList<Integer>(), results);
       return results;
   }
   
   private void search(int n, int k, int index, List<Integer> comb, List<List<Integer>> results) {
       if (comb.size() == k) {
           results.add(new ArrayList<Integer>(comb));
           return;
       }
       
       for (int i = index; i <= n; i++) {
           comb.add(i);
           search(n, k, i + 1, comb, results);
           comb.remove(comb.size() - 1);
       }
   }
}

标签：combination,int,ArrayList,List,results,Combinations,new
来源： https://www.cnblogs.com/blackraven25/p/15110040.html