8.5 Combinations

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

Approach I: Recursive, DFS

I thinks the code is very similar to that of Permutations. We also add "item" into result when the item.size() == k (or equal to num.length). 

The 1st difference is that we do not need to create a boolean array to denote whether an element has been used. 

Second, we use a "start" pointer here, and use helper(n, k, i+1, item, result), because we need to move forward one each time. 

{1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4} , {3, 4}. There is no combination in reverse order such as {2, 1}.

public class Solution {    public ArrayList<ArrayList<Integer>> combine(int n, int k){        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();        if(n < k) return result;        ArrayList<Integer> item = new ArrayList<Integer>();        dfs(n, k, 1, item, result);        return result;    }        public void dfs(int n, int k, int start, ArrayList<Integer> item, ArrayList<ArrayList<Integer>> result){        if(item.size() == k){            result.add(new ArrayList<Integer>(item));            return;        }        for(int i = start; i <=n; i++){            item.add(i);            dfs(n, k, i+1, item, result);            item.remove(item.size()-1);        }    }}

Note: http://blog.csdn.net/u010500263/article/details/18435495 has a very good explanation. 

"Parameters we need for a recursive call are the current combination, the combination result set, the n, the k, and the start point of iteration (e.g. if we chose 1 for the first position, start point of 2nd position is 2; while if we chose 2 for the first position, start point of second position is 3).  "

Approach II: DFS, easier to understand

Idea: First get combine (n-1, k-1), then randomly pick a number and add to it 

public class Solution {    public ArrayList<ArrayList<Integer>> combine(int n, int k) {        //This is similar to CC150 (9.4): to return all subsets of a set, but I didn't remember how to do it.        //The difference is that in this question, k = 2        //so instead of getting all subsets, we only get those with 2 elements.        //http://discuss.leetcode.com/questions/252/combinations        if(n < k) return null;        ArrayList<ArrayList<Integer>> lists = new ArrayList<ArrayList<Integer>>();        ArrayList<Integer> list;        if(k == 1){            for(int i = 1; i <=n; i++){                list = new ArrayList<Integer>();                list.add(i);                lists.add(list);            }            return lists;        }        //k>=2        for(int i = n; i >=k; i--){//i>=k, not i>=1            for(ArrayList<Integer> oneList: combine(i-1, k-1)){                oneList.add(i);                lists.add(oneList);            }        }        return lists;    }}

相关题目：

Subsets

本题可以看成是Subsets的特殊情况。当且仅当item.size() == k时，把item加入结果。

            result.add(new ArrayList<Integer>(item));