Combinations

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

Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.

For example,
If n = 4 and k = 2, a solution is:

[
  [2,4],
  [3,4],
  [2,3],
  [1,2],
  [1,3],
  [1,4],
]

解题思路：

最基本的组合题目，DFS+回溯，遇到了去重，解决方法也比较基本。现在写起来轻松多了。

public class Solution {
    public List<List<Integer>> combine(int n, int k) {
        List<List<Integer>> result = new LinkedList<List<Integer>>();
        List<Integer> current = new LinkedList<Integer>();
        int[] dict = new int[n];
        for(int i = 0; i < n; i++){
            dict[i] = i + 1;
        }
        dfs(result, current, dict, k, 0);
        return result;
    }
    
    public void dfs(List<List<Integer>> result, List<Integer> current, int[] dict, int k, int step){
        if(current.size() == k){
            result.add(new LinkedList(current));
            return;
        }
        
        for(int i = step; i < dict.length; i++){
            current.add(dict[i]);
            dfs(result, current, dict, k, i + 1);
            current.remove(current.size() - 1);
        }
    }
}

 注意是i+1，不是step+1。

 5个参数的解法

 

public class Solution {
    public List<List<Integer>> combine(int n, int k) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if(n == 0) {
            return res;
        }
        dfs(n, k, res, new ArrayList<Integer>(), 1);
        return res;
    }
    
    public void dfs(int n, int k, List<List<Integer>> res, List<Integer> cur, int start) {
        if(cur.size() == k) {
            res.add(new ArrayList(cur));
            return;
        }
        for(int i = start; i <= n; i++) {
            cur.add(i);
            dfs(n, k, res, cur, i + 1);
            cur.remove(cur.size() - 1);
        }
    }
}